Chapter 10 Circles (Additional Questions)

In geometry, a tangent to a circle is defined as a straight line that touches the circle at a single point. This point is called the point of tangency or the point of contact.

---

Considering the given options based on this definition:

(A) One point: This aligns directly with the definition of a tangent line, which intersects the circle at exactly one point.

(B) Two points: A line that intersects a circle at two distinct points is called a secant line.

(C) No points: A line that does not intersect the circle at all is a line that lies entirely outside the circle.

(D) Infinitely many points: A line cannot intersect a circle at infinitely many points unless the 'line' is actually part of the circle itself, which contradicts the definition of a straight line intersecting a curve.

---

Therefore, by definition, a tangent to a circle intersects the circle in one point.

---

The correct option is (A) One point.

In geometry, a secant to a circle is defined as a straight line that passes through the circle and intersects it at two distinct points.

---

Considering the given options based on this definition:

(A) One point: A line that intersects a circle at exactly one point is called a tangent line.

(B) Two points: This aligns directly with the definition of a secant line, which intersects the circle at exactly two distinct points.

(C) No points: A line that does not intersect the circle at all lies entirely outside the circle.

(D) Infinitely many points: A straight line can only intersect a circle at a maximum of two points. Intersection at infinitely many points would imply the line segment is part of the circle itself, which is not the definition of a secant.

---

Therefore, by definition, a secant to a circle intersects the circle in two points.

---

The correct option is (B) Two points.

A tangent is a line that intersects a circle at exactly one point, called the point of tangency.

---

A circle is a continuous curve made up of an infinite number of points on its circumference.

---

For each point on the circumference of the circle, there exists a unique tangent line that passes through that point.

---

Since there are infinitely many points on the circumference of a circle, there can be infinitely many tangent lines drawn to the circle.

---

The number of tangents from an external point is indeed limited (at most two), but the question asks about the total number of tangents a circle can have anywhere on its circumference.

---

Therefore, a circle can have infinitely many tangents.

---

The correct option is (D) Infinitely many.

This question refers to a fundamental theorem regarding tangents to a circle.

---

The theorem states that the tangent at any point on the circumference of a circle is perpendicular to the radius drawn to the point of contact.

---

Let O be the center of the circle, and let P be a point on the circle. If AB is the tangent line at point P, then the radius OP is perpendicular to the line AB.

---

This means the angle formed between the radius OP and the tangent line AB at the point of contact P is always $90^\circ$.

---

Comparing this with the given options:

(A) Parallel: This is incorrect. Parallel lines do not intersect.

(B) Perpendicular: This matches the theorem. The tangent is perpendicular to the radius at the point of contact.

(C) Equal: This term is not used to describe the geometric relationship between a line and a radius in this context.

(D) Intersecting at $45^\circ$: This is incorrect. The intersection angle is $90^\circ$, not $45^\circ$.

---

Therefore, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

---

The correct option is (B) Perpendicular.

Given:

Center of the circle = O

Radius of the circle, $r$ = OT = 5 cm (Let T be the point of tangency)

Distance of external point P from the center O, OP = 13 cm

PT is the tangent from external point P to the circle at point T.

---

To Find:

The length of the tangent PT.

---

Solution:

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

---

Therefore, the radius OT is perpendicular to the tangent PT at the point of tangency T.

---

Thus, triangle OTP is a right-angled triangle at T.

---

By applying the Pythagorean theorem in right-angled triangle OTP, we have:

---

Substitute the given values:

$(13)^2 = (5)^2 + \text{PT}^2$

$169 = 25 + \text{PT}^2$

---

Rearrange the equation to solve for PT$^2$:

$\text{PT}^2 = 169 - 25$

$\text{PT}^2 = 144$

---

Take the square root of both sides to find PT:

$\text{PT} = \sqrt{144}$

$\text{PT} = 12$

---

Since length must be a positive value, the length of the tangent PT is 12 cm.

---

The correct option is (B) 12 cm.

This question is based on a key theorem regarding tangents drawn from an external point to a circle.

---

The theorem states that the lengths of the two tangents drawn from an external point to a circle are equal.

---

Let P be an external point and let PA and PB be two tangents drawn from P to a circle with center O, touching the circle at points A and B respectively. The theorem states that the length of segment PA is equal to the length of segment PB.

---

Comparing this with the given options:

(A) Unequal: This contradicts the theorem.

(B) Equal: This matches the theorem.

(C) Proportional: While lengths might be involved in proportional relationships in other geometric contexts, the direct relationship between the two tangent lengths from the same external point is equality, not proportionality to each other.

(D) Perpendicular: This term describes angles, not lengths. Tangents are perpendicular to the radii at the point of contact, but the tangent lines themselves are not perpendicular to each other unless they meet at a specific angle (like $90^\circ$ if the quadrilateral formed by O, A, P, B is a square), and their lengths are still equal.

---

Therefore, the lengths of tangents drawn from an external point to a circle are equal.

---

The correct option is (B) Equal.

Let P be the external point from which two tangents PA and PB are drawn to a circle with center O. Let A and B be the points of contact on the circle.

---

We know that the radius through the point of contact is perpendicular to the tangent at that point.

---

Consider the quadrilateral PAOB. The sum of the interior angles of a quadrilateral is $360^\circ$.

So, in quadrilateral PAOB:

$\angle$APB + $\angle$PAO + $\angle$AOB + $\angle$OBP = $360^\circ$

---

Substitute the values of $\angle$PAO and $\angle$OBP:

$\angle$APB + $90^\circ$ + $\angle$AOB + $90^\circ$ = $360^\circ$

$\angle$APB + $\angle$AOB + $180^\circ$ = $360^\circ$

---

Rearrange the equation:

$\angle$APB + $\angle$AOB = $360^\circ - 180^\circ$

---

This relationship shows that the sum of the angle between the tangents ($\angle$APB) and the angle subtended by the line segment joining the points of contact at the center ($\angle$AOB) is $180^\circ$. Angles whose sum is $180^\circ$ are called supplementary angles.

---

Therefore, the angle between the tangents is supplementary to the angle subtended by the line segment joining the points of contact at the center.

---

Comparing this with the given options:

(A) Equal to: Incorrect.

(B) Supplementary to: Correct, as their sum is $180^\circ$.

(C) Complementary to: Incorrect, as complementary angles sum to $90^\circ$.

(D) Double: Incorrect.

---

The correct option is (B) Supplementary to.

Given:

P is an external point.

PA and PB are tangents from P to a circle with center O.

The angle between the tangents, $\angle$APB = $80^\circ$.

---

To Find:

The angle $\angle$POB.

---

Solution:

We know that the line segment joining the center to the external point P bisects the angle between the tangents.

---

Therefore, PO bisects $\angle$APB.

---

We also know that the radius is perpendicular to the tangent at the point of contact.

Thus, OB is perpendicular to the tangent PB.

---

Consider the right-angled triangle $\triangle$POB (right-angled at B).

The sum of angles in a triangle is $180^\circ$.

In $\triangle$POB:

$\angle$POB + $\angle$OBP + $\angle$BPO = $180^\circ$

---

Substitute the values from (i) and the perpendicularity property:

$\angle$POB + $90^\circ$ + $40^\circ$ = $180^\circ$

$\angle$POB + $130^\circ$ = $180^\circ$

---

Solve for $\angle$POB:

---

Therefore, the angle $\angle$POB is $50^\circ$.

---

The correct option is (D) $50^\circ$.

Given:

A quadrilateral ABCD is drawn to circumscribe a circle.

---

To Find:

The true statement among the given options.

---

Solution:

Let the circle touch the sides AB, BC, CD, and DA of the quadrilateral at points P, Q, R, and S respectively.

---

We know that the lengths of tangents drawn from an external point to a circle are equal.

Applying this property to the vertices of the quadrilateral (which are external points to the circle):

From vertex A, the tangents are AP and AS.

AP = AS

---

From vertex B, the tangents are BP and BQ.

BP = BQ

---

From vertex C, the tangents are CQ and CR.

CQ = CR

---

From vertex D, the tangents are DR and DS.

DR = DS

---

Now, consider the sum of opposite sides of the quadrilateral:

$AB + CD = (AP + PB) + (CR + RD)$

---

Substitute the equal tangent lengths into this sum:

$AB + CD = (AS + BQ) + (CQ + DS)$

$AB + CD = AS + BQ + CQ + DS$

---

Now consider the sum of the other pair of opposite sides:

$BC + DA = (BQ + QC) + (DS + SA)$

$BC + DA = BQ + QC + DS + SA$

---

Rearrange the terms in the second sum:

$BC + DA = AS + BQ + CQ + DS$

---

Comparing the expressions for $(AB + CD)$ and $(BC + DA)$, we see that they are equal.

Therefore,

AB + CD = BC + DA

---

This is a property of any quadrilateral that circumscribes a circle.

---

Comparing this result with the given options:

(A) $AB+BC = CD+DA$: Incorrect.

(B) $AB+CD = BC+DA$: Correct.

(C) $AB=CD, BC=DA$: This is only true for special cases, like a parallelogram circumscribing a circle (which must be a rhombus), but not for all circumscribing quadrilaterals.

(D) $AC = BD$: This relates to the diagonals and is not a general property.

---

The correct statement is $AB+CD = BC+DA$.

---

The correct option is (B) $AB+CD = BC+DA$.

Let the circle have its center at O and radius $r$.

---

Consider two parallel tangents to the circle. Let these tangents touch the circle at points A and B respectively.

---

We know that the radius drawn to the point of contact is perpendicular to the tangent at that point.

Therefore, the radius OA is perpendicular to the tangent at A, and the radius OB is perpendicular to the tangent at B.

---

Since the two tangents are parallel and OA and OB are both perpendicular to these parallel lines, the points A, O, and B must be collinear. This means that the line segment AB passes through the center O.

---

A line segment passing through the center of a circle and connecting two points on the circumference is called a diameter.

The length of the segment AB is the length of the diameter.

---

The length of the diameter is twice the radius.

Diameter = $2 \times \text{radius}$

AB = $2r$

---

The distance between two parallel lines is the length of the perpendicular segment between them. In this case, the segment AB is perpendicular to both parallel tangents and its length is $2r$.

---

Thus, the distance between the two parallel tangents is equal to the diameter of the circle, which is $2r$.

---

The correct option is (B) $2r$.

A tangent to a circle is a line that intersects the circle at exactly one point on its circumference.

---

Consider a point P that lies inside the circle.

---

Any straight line passing through a point inside the circle must necessarily intersect the circle at two distinct points as it enters and exits the circle. Such a line is a secant, not a tangent.

---

If a line from an internal point P were to touch the circle at only one point T, then the radius OT would be perpendicular to the line at T. However, the shortest distance from the center O to the line must be the radius $r$, and the distance from O to P is less than $r$ (since P is inside the circle). Any line through P would be further from O than P is, except at P itself. A tangent requires the shortest distance from O to the line to be exactly $r$. A line through an internal point cannot satisfy this condition.

---

Therefore, it is impossible to draw a line from a point inside the circle that touches the circle at only one point.

---

Thus, the number of tangents that can be drawn to a circle from a point inside the circle is zero.

---

Comparing this with the given options:

(A) One: Incorrect.

(B) Two: Incorrect. Two tangents can be drawn from an external point.

(C) Infinitely many: Incorrect.

(D) Zero: Correct.

---

The correct option is (D) Zero.

Given:

Distance of the external point P from the center O, OP = 25 cm.

Length of the tangent from P to the circle. Let the point of tangency be T. So, PT = 24 cm.

Let the radius of the circle be $r = \text{OT}$.

---

To Find:

The radius of the circle, $r$.

---

Solution:

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Thus, the radius OT is perpendicular to the tangent PT at point T.

---

This means that $\triangle$OTP is a right-angled triangle with the right angle at T.

---

In a right-angled triangle, we can apply the Pythagorean theorem, which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In $\triangle$OTP, the hypotenuse is OP (the side opposite the right angle).

---

Substitute the given values into the equation (i):

$(25)^2 = r^2 + (24)^2$

$625 = r^2 + 576$

---

Now, solve for $r^2$:

$r^2 = 625 - 576$

$r^2 = 49$

---

Take the square root of both sides to find the value of $r$:

$r = \sqrt{49}$

$r = 7$

---

Since the radius must be a positive value, the radius of the circle is 7 cm.

---

The correct option is (A) 7 cm.

Let's analyze the given Assertion (A) and Reason (R).

---

Assertion (A): The radius of a circle is perpendicular to the tangent at the point of contact.

This statement is a fundamental theorem in geometry related to circles and tangents. It states that if a line is tangent to a circle at a point, then the radius drawn from the center to that point of contact is perpendicular to the tangent line. This assertion is True.

---

Reason (R): This is a fundamental theorem in the study of tangents to a circle.

The statement in Reason (R) claims that the statement in Assertion (A) is a fundamental theorem. This is also True. The theorem described in A is indeed a cornerstone theorem when studying properties of tangents to circles.

---

Now, we need to determine if Reason (R) is the correct explanation for Assertion (A).

The reason explains *why* Assertion (A) is true in the context of geometry - because it is a well-established, fundamental theorem. The truth of Assertion (A) is derived directly from this established principle in the study of tangents.

---

Therefore, Reason (R) correctly explains why Assertion (A) is true.

---

Based on this analysis:

Both Assertion (A) and Reason (R) are true, and Reason (R) provides the correct explanation for Assertion (A).

---

The correct option is (A) Both A and R are true and R is the correct explanation of A.

Let's analyze the given Assertion (A) and Reason (R).

---

Assertion (A): From an external point, two tangents can be drawn to a circle.

This statement describes the maximum number of tangents that can be drawn to a circle from a point outside the circle. This is a well-established geometric property. Therefore, Assertion (A) is True.

---

Reason (R): The lengths of tangents drawn from an external point to a circle are equal.

This statement describes a property of the tangents drawn from an external point. It is a fundamental theorem stating that if two tangents are drawn from the same external point to a circle, their lengths from the external point to the points of contact are equal. Therefore, Reason (R) is also True.

---

Now, let's determine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) is about the *existence* and *number* of tangents from an external point (exactly two). Reason (R) is about a *property* of these tangents (their equal lengths).

The fact that the lengths are equal is a consequence of the tangents existing and being drawn from the same point. It does not explain *why* exactly two tangents can be drawn. The existence of two tangents from an external point is a different geometric concept, often related to the perpendicular bisector of the line segment joining the center and the external point, or construction using circles.

Therefore, while both statements are true, Reason (R) does not explain *why* Assertion (A) is true. It describes a property *of* those tangents.

---

Based on this analysis:

Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A).

---

The correct option is (B) Both A and R are true but R is not the correct explanation of A.

Let's match each term in Column A with its correct description in Column B:

---

(i) Tangent: A tangent to a circle is a line that intersects the circle at exactly one point. This matches description (c) Line intersecting the circle at exactly one point.

(i) $\to$ (c)

---

(ii) Secant: A secant to a circle is a line that intersects the circle at two distinct points. This matches description (b) Line intersecting the circle at two distinct points.

(ii) $\to$ (b)

---

(iii) Radius: A radius of a circle is a line segment joining the center of the circle to any point on its circumference. This matches description (a) Line segment joining center to circumference.

(iii) $\to$ (a)

---

(iv) Chord: A chord of a circle is a line segment joining any two points on the circumference of the circle. This matches description (d) Line segment joining two points on the circle.

(iv) $\to$ (d)

---

Combining the matches, we get:

(i) - (c)

(ii) - (b)

(iii) - (a)

(iv) - (d)

---

Now let's check the given options to find the correct matching:

(A) (i)-(c), (ii)-(b), (iii)-(a), (iv)-(d)

(B) (i)-(c), (ii)-(a), (iii)-(b), (iv)-(d)

(C) (i)-(a), (ii)-(b), (iii)-(c), (iv)-(d)

(D) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b)

---

Option (A) matches our correct pairing.

---

The correct option is (A) (i)-(c), (ii)-(b), (iii)-(a), (iv)-(d).

Given:

Center of the circular park = O

Radius of the park, $r = 8$ metres.

Distance of point B from the center O, OB = 17 metres.

The walkway is a tangent to the park's boundary at point A.

---

To Find:

The question asks for the length of the tangent segment from B to the walkway. Given the context of standard circle theorems and the provided options, this is interpreted as finding the length of a tangent segment drawn from the external point B to the circle.

---

Solution:

Let T be the point of contact of a tangent drawn from the external point B to the circle.

The radius of the circle is OT = $r = 8$ m.

The distance of point B from the center is OB = 17 m.

---

According to the theorem, the radius through the point of contact is perpendicular to the tangent at that point.

So, the radius OT is perpendicular to the tangent segment BT at point T.

---

Therefore, $\triangle$OTB is a right-angled triangle with the right angle at T.

In a right-angled triangle, we can use the Pythagorean theorem, which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In $\triangle$OTB, the hypotenuse is the side opposite the right angle, which is OB.

---

Substitute the given values into the equation (i):

$(17)^2 = (8)^2 + \text{BT}^2$

$289 = 64 + \text{BT}^2$

---

Solve for BT$^2$:

$\text{BT}^2 = 289 - 64$

$\text{BT}^2 = 225$

---

Take the square root of both sides to find the length of BT:

$\text{BT} = \sqrt{225}$

$\text{BT} = 15$

---

The length of the tangent segment from B to the circle is 15 metres.

Now, let's look at the options. Option (B) provides the expression $\sqrt{17^2 - 8^2}$, which is the intermediate step we used to find BT.

---

The correct option that represents the length of the tangent segment is $\sqrt{17^2 - 8^2}$ metres.

---

The correct option is (B) $\sqrt{17^2 - 8^2}$ metres.

Given:

A circular park with center O.

A walkway that is tangent to the circle at point A.

OA is the radius of the circle drawn to the point of tangency A.

---

To Find:

The angle between the radius OA and the walkway at point A.

---

Solution:

This question directly applies a fundamental theorem about tangents to a circle.

---

The theorem states that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

---

In this case, the walkway is the tangent line and point A is the point of contact. OA is the radius drawn from the center O to the point of contact A.

---

According to the theorem, the radius OA is perpendicular to the walkway (tangent) at point A.

Perpendicular lines intersect at an angle of $90^\circ$.

---

Therefore, the angle between the radius OA and the walkway at point A is $90^\circ$.

---

Comparing this with the given options:

(A) $0^\circ$: Incorrect.

(B) $45^\circ$: Incorrect.

(C) $90^\circ$: Correct.

(D) $180^\circ$: Incorrect.

---

The correct option is (C) $90^\circ$.

Given:

Two tangents are drawn from an external point P to a circle. Let the points of contact be A and B.

The angle between the tangents is $\angle$APB = $60^\circ$.

---

To Determine:

The type of triangle formed by the external point and the points of contact, which is $\triangle$PAB.

---

Solution:

We know that the lengths of tangents drawn from an external point to a circle are equal.

Therefore, the length of tangent PA is equal to the length of tangent PB.

---

Since two sides of $\triangle$PAB are equal (PA = PB), the triangle $\triangle$PAB is an isosceles triangle.

---

In an isosceles triangle, the angles opposite the equal sides are also equal. The angles opposite sides PA and PB are $\angle$PBA and $\angle$PAB respectively.

---

The sum of angles in a triangle is $180^\circ$. In $\triangle$PAB:

$\angle$APB + $\angle$PAB + $\angle$PBA = $180^\circ$

---

Substitute the given value $\angle$APB = $60^\circ$ and the equality $\angle$PAB = $\angle$PBA:

$60^\circ + \angle$PAB + $\angle$PAB = $180^\circ$

$60^\circ + 2 \angle$PAB = $180^\circ$

---

Solve for $\angle$PAB:

$2 \angle$PAB = $180^\circ - 60^\circ$

$2 \angle$PAB = $120^\circ$

---

Since $\angle$PAB = $\angle$PBA, we also have $\angle$PBA = $60^\circ$.

---

In $\triangle$PAB, we have:

$\angle$APB = $60^\circ$

$\angle$PAB = $60^\circ$

$\angle$PBA = $60^\circ$

---

A triangle with all three angles equal to $60^\circ$ is an equilateral triangle. In an equilateral triangle, all sides are also equal, so PA = PB = AB.

---

Therefore, the triangle formed by the external point and the points of contact is an equilateral triangle.

---

Comparing this with the given options:

(A) Isosceles right-angled: Incorrect. Angles are $60^\circ$, not $90^\circ$.

(B) Equilateral: Correct, as all angles are $60^\circ>.

(C) Scalene: Incorrect, as at least two sides (PA and PB) are equal.

(D) Isosceles but not equilateral: Incorrect, it is equilateral because all angles are $60^\circ>.

---

The correct option is (B) Equilateral.

A tangent is defined as a straight line that touches a circle at exactly one point.

---

The point where the tangent line and the circle meet is the single point that lies on both the line and the circle's circumference.

---

This specific common point is known by a particular term in geometry.

---

Let's consider the given options:

(A) Center: The center is inside the circle, not on the circumference where the tangent touches.

(B) Point of intersection: While it is an intersection point, this is a general term. A more precise term exists for this specific type of intersection involving a tangent.

(C) Point of contact: This is the standard and precise geometric term for the single point where a tangent line touches a curve, such as a circle.

(D) Vertex: A vertex is typically associated with angles, polygons, or polyhedra, not the intersection point of a tangent and a circle.

---

Therefore, the common point of a tangent to a circle and the circle is called the point of contact or point of tangency.

---

The correct option is (C) Point of contact.

Let's analyze each statement regarding a tangent to a circle.

---

(A) It is a chord.

A chord is a line segment connecting two distinct points on the circle. A tangent is a line (which extends infinitely in both directions) that intersects the circle at only one point. Therefore, a tangent is not a chord. This statement is False.

---

(B) It is a line passing through the center.

A line passing through the center of a circle intersects the circle at two points (unless the circle's radius is zero, which is not the usual case). A tangent intersects the circle at exactly one point. Therefore, a tangent cannot pass through the center of the circle. This statement is False.

---

(C) It intersects the circle at exactly one point.

This statement is the fundamental definition of a tangent line to a circle. A tangent line is characterized by meeting the circle at precisely one point, called the point of tangency. This statement is True.

---

(D) It is always perpendicular to a diameter.

A tangent is perpendicular to the radius at the point of contact. A diameter is a line segment passing through the center and having its endpoints on the circle; it contains a radius. A tangent is perpendicular to the *unique* diameter that passes through the point of contact. However, it is not perpendicular to *every* diameter. For example, a tangent line is parallel to the diameters that are perpendicular to the radius at the point of contact. While there is a perpendicular relationship involving a diameter, the statement "always perpendicular to a diameter" can be misleading and (C) is a more universally applicable definition. Let's clarify: The tangent at point P is perpendicular to the diameter that passes through P. This statement is True in the sense that there is *a* diameter it's perpendicular to, but the phrasing could be interpreted as *all* diameters, which is false. Compared to the precise definition in (C), (C) is a more direct and accurate description of *what* a tangent is.

---

Considering the options, the statement that is unequivocally and fundamentally TRUE about a tangent to a circle, serving as its definition, is that it intersects the circle at exactly one point.

---

The correct option is (C) It intersects the circle at exactly one point.

Let's define the terms given in the options in relation to a circle:

---

(A) Chord: A chord is a line segment that connects two distinct points on the circumference of a circle.

(B) Tangent: A tangent is a line that intersects the circle at exactly one point (the point of contact).

(C) Secant: A secant is a line that intersects the circle at exactly two distinct points.

(D) Diameter: A diameter is a special type of chord that passes through the center of the circle. It is also a line segment.

---

The question asks for the name of a line that intersects the circle in two distinct points.

---

Based on the definitions above, the term that fits this description is a secant.

---

The correct option is (C) Secant.

Given:

A circle and a secant line intersecting the circle at two distinct points.

---

To Find:

The number of tangents that can be drawn to the circle which are parallel to the given secant.

---

Solution:

Let the given secant line be L, intersecting the circle at points A and B. Let O be the center of the circle.

---

We know that a radius is perpendicular to a tangent at the point of contact. If a tangent is to be parallel to the secant L, then the radius drawn to the point of tangency must be perpendicular to both the tangent and the secant L.

---

Consider a line M passing through the center O and perpendicular to the given secant L. This line M will intersect the circle at two points. Let these points be P and Q.

---

At each of these two points P and Q, we can draw a unique tangent line. Let the tangent at P be $T_1$ and the tangent at Q be $T_2$.

---

Since OP is the radius through the point of contact P, OP is perpendicular to the tangent $T_1$. We constructed line M (containing OP) to be perpendicular to the secant L. Since both $T_1$ and L are perpendicular to the same line M, $T_1$ must be parallel to L.

Similarly, since OQ is the radius through the point of contact Q, OQ is perpendicular to the tangent $T_2$. Line M (containing OQ) is perpendicular to the secant L. Since both $T_2$ and L are perpendicular to the same line M, $T_2$ must be parallel to L.

---

The points P and Q are the only two points on the circle where the radius (or diameter PQ) is perpendicular to the secant L. Therefore, these are the only two points at which a tangent can be drawn parallel to the secant L.

---

Thus, exactly two tangents can be drawn to a circle parallel to a given secant.

---

Comparing this with the given options:

(A) One: Incorrect.

(B) Two: Correct.

(C) Infinitely many: Incorrect.

(D) Zero: Incorrect.

---

The correct option is (B) Two.

Let P be the external point from which two tangents are drawn to a circle with center O. Let the points where the tangents touch the circle be A and B.

---

We are considering the segment OP, which joins the center O to the external point P.

---

We can consider the two triangles formed, $\triangle$PAO and $\triangle$PBO.

---

We know the following properties:

1. The lengths of tangents drawn from an external point to a circle are equal.

---

2. The radius to the point of contact is perpendicular to the tangent at that point.

---

Also, OA and OB are radii of the same circle, so:

---

OP is a common side to both triangles $\triangle$PAO and $\triangle$PBO.

---

Consider $\triangle$PAO and $\triangle$PBO. We have OA = OB, OP is common, and PA = PB. By the SSS congruence criterion, $\triangle$PAO $\cong$ $\triangle$PBO. Alternatively, using the right angles at A and B, we have OA = OB (Side), OP = OP (Hypotenuse), and $\angle$OAP = $\angle$OBP = $90^\circ$ (Right angle). By the RHS congruence criterion, $\triangle$PAO $\cong$ $\triangle$PBO.

---

Since the triangles are congruent, their corresponding parts are equal (CPCTC).

Specifically, the angles opposite the common side OP are equal:

$\angle$APO = $\angle$BPO

---

The angle $\angle$APB is the angle between the two tangents PA and PB. The segment OP divides $\angle$APB into two equal angles, $\angle$APO and $\angle$BPO.

---

Therefore, the segment joining the center to the external point bisects the angle between the tangents.

---

As a side note, OP also bisects the angle between the radii drawn to the points of contact (i.e., $\angle$AOB), since $\angle$AOP = $\angle$BOP (CPCTC).

---

Looking at the options, the question asks what angle formed at P is bisected by OP. This is the angle between the tangent segments PA and PB, which are parts of the tangent lines.

---

The correct option is (C) Tangents.

This question describes a theorem in geometry known as the Alternate Segment Theorem.

---

The theorem states: The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

---

Let PT be the tangent at point P, and PQ be a chord through P.

The angle between the tangent and the chord can be $\angle$TPQ or $\angle$RPQ, where R is a point on the tangent on the other side of PQ.

---

The chord PQ divides the circle into two segments: a major segment and a minor segment.

The angles subtended by the chord PQ in these segments are angles formed by joining any point on the circumference within that segment to the endpoints of the chord P and Q.

---

The Alternate Segment Theorem specifically states that:

1. The angle between the tangent PT and the chord PQ ($\angle$TPQ) is equal to the angle subtended by the chord PQ in the segment that does not contain the angle $\angle$TPQ. This segment is called the alternate segment.

2. The angle between the tangent and the chord on the other side ($\angle$RPQ) is equal to the angle subtended by the chord PQ in the other segment (the one containing $\angle$TPQ).

---

The term "alternate segment" refers to the segment of the circle that is on the opposite side of the chord from the angle being considered.

---

Comparing this with the given options:

(A) Same: Incorrect. The angle is in the alternate segment.

(B) Alternate: Correct. This matches the theorem.

(C) Major: While the alternate segment might be the major segment in some cases (e.g., if $\angle$TPQ is acute), it's not always the major segment.

(D) Minor: While the alternate segment might be the minor segment in some cases (e.g., if $\angle$TPQ is obtuse), it's not always the minor segment.

---

The theorem specifically uses the term "alternate segment".

---

The correct option is (B) Alternate.

Let P be an external point from which two tangents PA and PB are drawn to a circle with center O, touching the circle at points A and B respectively.

---

Let's analyze each statement:

---

(A) Their lengths are equal.

This is a fundamental theorem regarding tangents from an external point. The lengths of the tangent segments from the external point to the points of contact are equal.

This statement is TRUE.

---

(B) They subtend equal angles at the center.

Consider the triangles $\triangle$PAO and $\triangle$PBO. We know that OA = OB (radii of the same circle), PA = PB (from statement A, which is true), and OP is common to both triangles. By the SSS congruence criterion, $\triangle$PAO $\cong$ $\triangle$PBO. Therefore, corresponding angles are equal.

These angles are subtended by the tangent segments (or the lines from the center to the points of contact) at the center O. Therefore, the tangents subtend equal angles at the center.

This statement is TRUE.

---

(C) They are perpendicular to the radius at the external point.

The theorem states that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

So, the tangent PA is perpendicular to the radius OA at point A ($\angle$OAP = $90^\circ$), and the tangent PB is perpendicular to the radius OB at point B ($\angle$OBP = $90^\circ$).

The external point is P. The radius from the center O to the external point P is the segment OP. The tangents PA and PB are lines passing through P, but they are not perpendicular to the segment OP at P (unless P is on the circle, which is not the case for an external point). In fact, OP lies along the angle bisector of $\angle$APB.

This statement is FALSE.

---

(D) The line segment joining the center to the external point bisects the angle between the tangents.

From the congruence of $\triangle$PAO and $\triangle$PBO (as shown in part B), the corresponding angles at point P are equal.

The angle between the tangents is $\angle$APB. The segment OP divides $\angle$APB into two angles, $\angle$APO and $\angle$BPO. Since these two angles are equal, the segment OP bisects the angle $\angle$APB.

This statement is TRUE.

---

Based on the analysis, the statements that are TRUE are (A), (B), and (D).

---

The correct options are (A), (B), and (D).

This question refers to the Tangent-Secant Theorem, which is a theorem relating a tangent line and a secant line drawn to a circle from a single external point.

---

Let P be an external point. Let PA be a tangent segment from P to the circle, where A is the point of tangency. Let PBC be a secant line from P that intersects the circle at points B and C, where B is between P and C.

---

The Tangent-Secant Theorem states that the square of the length of the tangent segment from the external point is equal to the product of the lengths of the secant segment from the external point and its external part.

---

In this case:

The tangent segment is PA.

The secant segment from P is PC (the entire segment from P to the far intersection point C).

The external part of the secant segment is PB (the segment from P to the near intersection point B).

---

According to the theorem, the relationship is:

---

Comparing this formula with the given options:

(A) $PB \times PC$: This matches the formula derived from the Tangent-Secant Theorem.

(B) $PB + PC$: This is a sum, not a product, and does not match the theorem.

(C) $PB^2 + PC^2$: This involves squares and addition, related to Pythagorean theorem or distance formulas, not the Tangent-Secant Theorem.

(D) $PB \times BC$: BC is the segment of the secant inside the circle. The theorem uses the external segment (PB) and the whole secant segment (PC = PB + BC), not the internal segment (BC).

---

Therefore, the correct completion of the formula is $PA^2 = PB \times PC$.

---

The correct option is (A) $PB \times PC$.

Consider two circles with centers $O_1$ and $O_2$, and radii $r_1$ and $r_2$ respectively. Assume, without loss of generality, that the circle with center $O_1$ is the larger circle, so $r_1 > r_2$.

---

When two circles touch internally, they have exactly one common point. This common point is called the point of contact.

---

A key property of circles touching internally is that their centers ($O_1$ and $O_2$) and the point of contact (let's call it T) are collinear. This means they lie on the same straight line.

---

For internal touching, the point of contact T lies on the line segment extending from the smaller circle's center $O_2$ through its circumference to the larger circle's circumference. T also lies on the line extending from the larger circle's center $O_1$ through its circumference.

---

Specifically, the center of the smaller circle ($O_2$) lies between the center of the larger circle ($O_1$) and the point of contact (T).

---

The distance from $O_1$ to T is the radius of the larger circle, $r_1$. So, $O_1\text{T} = r_1$.

The distance from $O_2$ to T is the radius of the smaller circle, $r_2$. So, $O_2\text{T} = r_2$.

---

Since $O_1$, $O_2$, and T are collinear and $O_2$ is between $O_1$ and T, the distance from $O_1$ to T is the sum of the distance from $O_1$ to $O_2$ and the distance from $O_2$ to T.

---

Substitute the radii values:

$r_1 = O_1O_2 + r_2$

---

Now, solve for the distance between the centers, $O_1O_2$:

---

The distance between the centers is the difference between the radii of the two circles.

---

Comparing this with the given options:

(A) Sum: Incorrect. The sum of radii ($r_1 + r_2$) is the distance between centers when circles touch externally.

(B) Difference: Correct. The distance is the difference between the radii.

(C) Product: Incorrect.

(D) Ratio: Incorrect.

---

The correct option is (B) Difference.

Consider two circles with centers $O_1$ and $O_2$, and radii $r_1$ and $r_2$ respectively.

---

When two circles touch externally, they have exactly one common point, the point of contact. This point lies on the line segment joining the centers of the two circles.

---

Let T be the point of contact where the two circles touch externally.

A key property of circles touching externally is that their centers ($O_1$ and $O_2$) and the point of contact (T) are collinear, and the point of contact T lies between the two centers $O_1$ and $O_2$.

---

The distance from $O_1$ to the point of contact T is the radius of the first circle, $r_1$. So, $O_1\text{T} = r_1$.

The distance from $O_2$ to the point of contact T is the radius of the second circle, $r_2$. So, $O_2\text{T} = r_2$.

---

Since $O_1$, T, and $O_2$ are collinear and T is between $O_1$ and $O_2$, the distance between the centers ($O_1O_2$) is the sum of the distances $O_1\text{T}$ and $O_2\text{T}$.

---

Substitute the radii values:

---

Thus, the distance between the centers of two circles that touch externally is equal to the sum of their radii.

---

Comparing this with the given options:

(A) Sum: Correct. The distance is the sum of the radii.

(B) Difference: Incorrect. The difference of radii is the distance between centers when circles touch internally.

(C) Product: Incorrect.

(D) Ratio: Incorrect.

---

The correct option is (A) Sum.

Given:

Circle with center O.

Tangent line AB touching the circle at point P.

OP is the radius drawn to the point of contact P.

Radius of the circle = 7 cm (This information is relevant for other potential questions but not directly needed to find the angle).

---

To Find:

The angle between the radius OP and the tangent line AB at the point of contact P, i.e., $\angle$OPA.

---

Solution:

This question is based on a fundamental geometric theorem concerning tangents to a circle.

---

The theorem states that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

---

In this case, AB is the tangent line and P is the point of contact. OP is the radius drawn from the center O to the point of contact P.

---

According to the theorem, the radius OP is perpendicular to the tangent AB at point P.

Lines that are perpendicular intersect at an angle of $90^\circ$.

---

Therefore, the angle between the radius OP and the tangent AB at point P is $90^\circ$.

---

Comparing this result with the given options:

(A) $0^\circ$: Incorrect.

(B) $45^\circ$: Incorrect.

(C) $90^\circ$: Correct.

(D) $180^\circ$: Incorrect.

---

The correct option is (C) $90^\circ$.

Given:

Circle with center O.

Secant line CD intersecting the circle at points Q and R.

Distance from the center O to the chord QR = 5 cm.

Radius of the circle, OR = OQ = 7 cm.

---

To Find:

The length of the chord QR.

---

Solution:

Draw a perpendicular from the center O to the chord QR. Let the point where the perpendicular intersects QR be M.

The distance from the center O to the chord QR is OM = 5 cm.

---

We know a fundamental theorem: The perpendicular from the center of a circle to a chord bisects the chord.

Therefore, M is the midpoint of the chord QR.

---

Consider the triangle OMR. Since OM is perpendicular to QR, $\triangle$OMR is a right-angled triangle at M.

---

In right-angled $\triangle$OMR, we can apply the Pythagorean theorem. The hypotenuse is the radius OR (opposite the right angle).

---

Substitute the given values:

$(7)^2 = (5)^2 + \text{MR}^2$

$49 = 25 + \text{MR}^2$

---

Solve for MR$^2$:

$\text{MR}^2 = 49 - 25$

$\text{MR}^2 = 24$

---

Take the square root of both sides to find the length of MR:

$\text{MR} = \sqrt{24}$ cm

---

Since M is the midpoint of QR, the length of the chord QR is twice the length of MR.

QR = $2 \times \text{MR}$

QR = $2 \times \sqrt{24}$ cm

---

Comparing this result with the given options:

(A) 14 cm: This would be the diameter ($2 \times 7$), which is the length of the chord only if it passes through the center (distance = 0). Incorrect.

(B) $\sqrt{7^2 - 5^2} = \sqrt{24}$ cm: This is the length of half the chord (MR). Incorrect.

(C) $2 \times \sqrt{7^2 - 5^2} = 2\sqrt{24}$ cm: This matches our calculated length of the chord QR. Correct.

(D) $7 - 5 = 2$ cm: Incorrect.

---

The correct option is (C) $2 \times \sqrt{7^2 - 5^2} = 2\sqrt{24}$ cm.

Given:

A circle with center O and radius $r$.

A tangent is drawn to the circle at point P.

---

To Find:

The distance of the tangent line from the center O.

---

Solution:

The distance from a point (in this case, the center O) to a line (in this case, the tangent line) is defined as the length of the perpendicular segment from the point to the line.

---

Let the tangent line be denoted by $t$. Point P is the point of contact on the circle where the tangent touches it.

---

We know the fundamental theorem regarding tangents: The radius drawn to the point of contact is perpendicular to the tangent at that point.

---

The segment OP is the radius drawn from the center O to the point of contact P.

---

According to the theorem, the radius OP is perpendicular to the tangent line $t$ at the point P.

This means the segment OP itself is the perpendicular segment from the center O to the tangent line $t$.

---

Therefore, the distance from the center O to the tangent line is the length of the segment OP.

---

Comparing this result with the given options:

(A) $r$: Correct. The distance is equal to the radius.

(B) $2r$: This is the diameter. Incorrect.

(C) $r/2$: Half the radius. Incorrect.

(D) 0: This would mean the tangent passes through the center, which is not possible for a tangent. Incorrect.

---

The correct option is (A) $r$.

Given:

Two disjoint circles.

---

To Find:

The number of common tangents that can be drawn to these two circles.

---

Solution:

Two circles are said to be disjoint if they do not intersect or touch each other at any point. This means the distance between their centers is greater than the sum of their radii.

---

A common tangent to two circles is a line that is tangent to both circles.

Common tangents can be classified into two types:

1. Direct (or Transverse) Common Tangents: These are common tangents such that the two circles lie on the same side of the tangent line.

2. Transverse (or Direct) Common Tangents: These are common tangents such that the two circles lie on opposite sides of the tangent line.

---

For two disjoint circles, we can draw both types of common tangents.

---

We can draw two direct common tangents. These tangents will intersect at a point on the line joining the centers of the circles, but outside the segment between the centers.

---

We can also draw two transverse common tangents. These tangents will intersect at a point on the line segment joining the centers of the circles.

---

Since there are two direct common tangents and two transverse common tangents for two disjoint circles, the total number of common tangents is the sum of these.

Total common tangents = Number of direct common tangents + Number of transverse common tangents

Total common tangents = 2 + 2

Total common tangents = 4

---

Therefore, four common tangents can be drawn to two disjoint circles.

---

Comparing this with the given options:

(A) 1: Incorrect.

(B) 2: Incorrect.

(C) 3: Incorrect. This is the number of common tangents if the circles touch externally.

(D) 4: Correct.

---

The correct option is (D) 4.

Given:

PA and PB are tangents from external point P to a circle with center O.

The angle between the tangents, $\angle$APB = $50^\circ$.

---

To Find:

The angle $\angle$AOB.

---

Solution:

We know that the radius through the point of contact is perpendicular to the tangent at that point.

Therefore, OA is perpendicular to tangent PA, and OB is perpendicular to tangent PB.

---

Consider the quadrilateral PAOB. The sum of the interior angles of a quadrilateral is $360^\circ$.

In quadrilateral PAOB:

$\angle$APB + $\angle$PAO + $\angle$AOB + $\angle$OBP = $360^\circ$

---

Substitute the known values:

$50^\circ + 90^\circ + \angle$AOB + $90^\circ$ = $360^\circ$

$230^\circ + \angle$AOB = $360^\circ$

---

Solve for $\angle$AOB:

---

Therefore, the angle $\angle$AOB is $130^\circ$.

---

As an alternative, we know that the angle between the tangents from an external point is supplementary to the angle subtended by the segment joining the points of contact at the center (from Question 7).

So, $\angle$APB + $\angle$AOB = $180^\circ$

$50^\circ + \angle$AOB = $180^\circ$

$\angle$AOB = $180^\circ - 50^\circ$

$\angle$AOB = $130^\circ$

---

Both methods yield the same result.

---

Comparing this with the given options:

(A) $50^\circ$: Incorrect.

(B) $100^\circ$: Incorrect.

(C) $130^\circ$: Correct.

(D) $80^\circ$: Incorrect.

---

The correct option is (C) $130^\circ$.

Given:

A triangle ABC.

A circle is inscribed inside the triangle, touching the sides AB, BC, and CA.

The points of tangency are P on AB, Q on BC, and R on CA.

---

To Find:

Identify the true statement among the given options based on the properties of tangents drawn from external points to a circle.

---

Solution:

The vertices of the triangle A, B, and C are external points to the inscribed circle.

---

We know a key theorem: The lengths of tangents drawn from an external point to a circle are equal.

---

Let's apply this theorem to each vertex:

From external point A, the tangent segments to the circle are AP (on side AB) and AR (on side CA).

---

From external point B, the tangent segments to the circle are BP (on side AB) and BQ (on side BC).

---

From external point C, the tangent segments to the circle are CQ (on side BC) and CR (on side CA).

---

Now, let's examine the given options in light of these equalities:

(A) $AP = BP$: AP and BP are tangent segments from different external points (A and B). Their lengths are not necessarily equal in a general triangle.

(B) $BQ = CQ$: BQ and CQ are tangent segments from different external points (B and C). Their lengths are not necessarily equal.

(C) $AR = CR$: AR and CR are tangent segments from different external points (A and C). Their lengths are not necessarily equal.

(D) $AP = AR$: AP and AR are tangent segments from the same external point A. According to the theorem, their lengths are equal.

---

Therefore, the statement that is TRUE based on tangent properties is $AP = AR$.

---

The correct option is (D) $AP = AR$.

Given:

Let O be the center of the circle. Let the two radii be OA and OB, where A and B are points on the circle.

The angle between the two radii is $\angle$AOB = $110^\circ$.

Let the tangents at points A and B intersect at an external point P.

---

To Find:

The angle between the tangents at the ends of the radii (at points A and B), which is $\angle$APB.

---

Solution:

We know that the radius at the point of contact is perpendicular to the tangent at that point.

Therefore, OA is perpendicular to the tangent PA, and OB is perpendicular to the tangent PB.

---

Consider the quadrilateral PAOB. The sum of the interior angles of a quadrilateral is $360^\circ$.

In quadrilateral PAOB:

$\angle$APB + $\angle$PAO + $\angle$AOB + $\angle$OBP = $360^\circ$

---

Substitute the known values:

$\angle$APB + $90^\circ$ + $110^\circ$ + $90^\circ$ = $360^\circ$

$\angle$APB + $290^\circ$ = $360^\circ$

---

Solve for $\angle$APB:

---

Therefore, the angle between the tangents at the ends of the radii is $70^\circ$.

---

Alternatively, we can use the theorem that the angle between two tangents drawn from an external point is supplementary to the angle subtended by the line segment joining the points of contact at the center.

The points of contact are A and B, and the angle subtended by the segment AB at the center O is $\angle$AOB = $110^\circ$. The angle between the tangents is $\angle$APB.

So, $\angle$APB + $\angle$AOB = $180^\circ$

$\angle$APB + $110^\circ = 180^\circ$

$\angle$APB = $180^\circ - 110^\circ$

$\angle$APB = $70^\circ$

---

Both methods confirm the result.

---

Comparing this with the given options:

(A) $110^\circ$: Incorrect.

(B) $70^\circ$: Correct.

(C) $55^\circ$: Incorrect.

(D) $90^\circ$: Incorrect.

---

The correct option is (B) $70^\circ$.

Given:

Equation of the circle is $x^2 + y^2 = r^2$, which is a circle centered at the origin (0, 0) with radius $r$.

A point on the circle is $(x_0, y_0)$.

---

To Find:

The equation of the tangent line to the circle at the given point $(x_0, y_0)$.

---

Solution:

For a circle with the equation $x^2 + y^2 = r^2$, the equation of the tangent line at a specific point $(x_0, y_0)$ that lies on the circle is given by a standard formula.

---

The standard formula for the tangent to $x^2 + y^2 = r^2$ at $(x_0, y_0)$ is obtained by replacing $x^2$ with $x \cdot x_0$ and $y^2$ with $y \cdot y_0$.

---

Thus, the equation of the tangent is:

---

Let's compare this with the given options:

(A) $xx_0 + yy_0 = r^2$: This matches the standard formula for the tangent at $(x_0, y_0)$.

(B) $x+y = r$: This is a linear equation, but not the correct formula for a tangent at an arbitrary point $(x_0, y_0)$.

(C) $x/x_0 + y/y_0 = 1$: This can be rearranged to $xy_0 + yx_0 = x_0y_0$, which is a linear equation, but not the correct tangent equation for $x^2+y^2=r^2$.

(D) $y - y_0 = m(x - x_0)$: This is the general point-slope form of a line passing through $(x_0, y_0)$. While the tangent is indeed a line of this form, this option does not specify the correct slope $m = -\frac{x_0}{y_0}$ (for $y_0 \neq 0$) which is required for it to be the tangent. It represents any line through $(x_0, y_0)$, not specifically the tangent.

---

The correct representation of the tangent to the circle $x^2+y^2=r^2$ at the point $(x_0, y_0)$ is given by the standard formula.

---

The correct option is (A) $xx_0 + yy_0 = r^2$.

Given:

Two circles touch externally.

---

To Find:

The number of common tangents that can be drawn to these two circles.

---

Solution:

When two circles touch externally, they have exactly one common point, which is the point of contact. The distance between their centers is equal to the sum of their radii.

---

A common tangent is a line that is tangent to both circles.

Common tangents can be categorized as:

1. Direct Common Tangents: The circles lie on the same side of the tangent. For circles touching externally, there are two direct common tangents.

2. Transverse Common Tangents: The circles lie on opposite sides of the tangent. For circles touching externally, there is one transverse common tangent which passes through the point of contact.

---

The total number of common tangents is the sum of the direct and transverse common tangents.

Total common tangents = Number of direct common tangents + Number of transverse common tangents

Total common tangents = 2 + 1

Total common tangents = 3

---

Therefore, three common tangents can be drawn to two circles that touch externally.

---

Comparing this with the given options:

(A) 1: Incorrect.

(B) 2: Incorrect. This is the number of direct tangents, but we also have a transverse one.

(C) 3: Correct.

(D) 4: Incorrect. This is the number for disjoint circles.

---

The correct option is (C) 3.

Let O be the center of a circle, and let $t$ be a tangent line to the circle at a point P.

---

The distance from a point (the center O) to a line (the tangent line $t$) is defined as the length of the perpendicular segment drawn from the point to the line.

---

According to a fundamental theorem related to tangents and circles, the radius drawn from the center to the point of contact is always perpendicular to the tangent line at that point.

---

In this case, OP is the radius drawn from the center O to the point of contact P on the tangent line $t$.

By the theorem, OP is perpendicular to the tangent line $t$ at P.

---

Since OP is the perpendicular segment from O to the line $t$, its length represents the distance from the center to the tangent line.

The length of the segment OP is the radius of the circle.

---

Therefore, the distance from the center of a circle to a tangent line is always equal to the radius of the circle.

---

Comparing this with the given options:

(A) Diameter: Incorrect. Diameter is twice the radius.

(B) Radius: Correct.

(C) Chord length: Incorrect. Chord length varies depending on the chord.

(D) Secant length: Incorrect. Secants are lines, not segments with a defined single length.

---

The correct option is (B) Radius.

Given:

Two circles that touch each other at a single point.

---

To Find:

The number of common tangents that can be drawn to these two circles.

---

Solution:

When two circles touch each other at a single point, there are two possible configurations:

1. Circles touch externally: The circles are outside each other, sharing one point of contact.

2. Circles touch internally: One circle is inside the other, sharing one point of contact on the circumference of the larger circle.

---

Let's analyze the number of common tangents for each case:

Case 1: Circles touch externally

In this case, there is one common tangent that passes through the point of contact (a transverse common tangent). Additionally, there are two common tangents that do not pass through the point of contact, with both circles lying on the same side of each tangent (direct common tangents).

Total common tangents = 1 (transverse) + 2 (direct) = 3.

---

Case 2: Circles touch internally

In this case, there is only one common tangent, which passes through the point of contact. Both circles lie on the same side of this tangent (it is a direct common tangent). There are no transverse common tangents.

Total common tangents = 1.

---

The question states that the circles touch at a single point and that "A common tangent line is drawn at this point". This confirms it is a touching scenario, and that the tangent at the point of contact exists (which is true for both internal and external touching). However, the question asks for the total number of common tangents that *can be drawn*.

---

Considering the options provided (1, 2, 3, 4), both 1 and 3 are possibilities depending on whether the touching is internal or external. In the context of problems asking for the number of common tangents based on relative positions, the case of external touching (3 tangents) is a distinct and commonly presented scenario.

Given the options and typical geometric problem settings, it is most likely that the question intends to refer to the number of tangents in the case where the circles, while touching at one point, allow for more than just that single tangent at the contact point itself.

---

Thus, the scenario allowing for 3 common tangents (external touching) aligns with one of the options.

---

The number of common tangents for two circles touching externally is 3.

---

The correct option is (C) 3.

Given:

An external point P.

A circle with center O and radius $r$.

Length of the tangent from P to the circle is $l$. Let the point of tangency be T, so PT = $l$.

Distance from the external point P to the center O is $d$, so OP = $d$.

---

To Find:

The radius $r$ of the circle in terms of $d$ and $l$.

---

Solution:

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Therefore, the radius OT is perpendicular to the tangent PT at the point of tangency T.

---

This means that $\triangle$OTP is a right-angled triangle with the right angle at T.

The sides of the triangle are OT (radius $r$), PT (tangent length $l$), and OP (distance from center to external point $d$).

---

In a right-angled triangle, we can apply the Pythagorean theorem, which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In $\triangle$OTP, the hypotenuse is OP (the side opposite the right angle at T).

---

Substitute the given variables into the equation:

$d^2 = r^2 + l^2$

---

We need to find the radius $r$. Let's rearrange the equation to solve for $r^2$:

$r^2 = d^2 - l^2$

---

To find $r$, take the square root of both sides:

$r = \sqrt{d^2 - l^2}$

Note that since $d$ is the hypotenuse in a right triangle where $l$ is a leg, it must be true that $d \ge l$. Since P is an external point, $d > r$, and $d > l$, so $d^2 - l^2 > 0$, ensuring the square root is real.

---

Comparing this derived formula with the given options:

(A) $r = d - l$: Incorrect.

(B) $r = d + l$: Incorrect.

(C) $r = \sqrt{d^2 - l^2}$: Correct.

(D) $r = \sqrt{l^2 - d^2}$: Incorrect. This would require $l \ge d$, which is not the case since $d$ is the hypotenuse.

---

The formula for the radius $r$ is $\sqrt{d^2 - l^2}$.

---

The correct option is (C) $r = \sqrt{d^2 - l^2}$.

A tangent to a circle is a straight line that touches the circle at only one point.

This unique point where the tangent touches the circle is known as the point of contact.

---

A circle can have infinitely many tangents.

This is because a tangent can be drawn at every point on the circumference of the circle, and there are infinitely many points on a circle's circumference.

Given:

Radius of the circle, $OP = 7 \text{ cm}$.

Distance of point Q from the center O, $OQ = 25 \text{ cm}$.

---

To Find:

The length of the tangent PQ.

---

Solution:

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Therefore, the angle between the radius OP and the tangent PQ at the point of contact P is $90^\circ$.

$\angle OPQ = 90^\circ$.

Thus, $\triangle OPQ$ is a right-angled triangle at P.

By applying the Pythagorean theorem in $\triangle OPQ$, we have:

$OP^2 + PQ^2 = OQ^2$

Substitute the given values into the equation:

$(7)^2 + PQ^2 = (25)^2$

$49 + PQ^2 = 625$

Now, isolate $PQ^2$ by subtracting 49 from both sides:

$PQ^2 = 625 - 49$

Performing the subtraction:

$\begin{array}{cc} & 6 & 2 & 5 \\ - & & 4 & 9 \\ \hline & 5 & 7 & 6 \\ \hline \end{array}$

So, $PQ^2 = 576$.

To find PQ, take the square root of both sides:

$PQ = \sqrt{576}$

The square root of 576 is 24.

$PQ = 24$

Since the given lengths are in cm, the length of PQ is $24 \text{ cm}$.

---

Answer:

The length of the tangent PQ is $24 \text{ cm}$.

Given:

Length of the tangent from point Q to the circle, $PQ = 20 \text{ cm}$.

Distance of point Q from the center O, $OQ = 29 \text{ cm}$.

---

To Find:

The radius of the circle, $OP$.

---

Solution:

Let O be the center of the circle and P be the point of contact of the tangent from Q.

We know that the radius through the point of contact is perpendicular to the tangent at that point.

Therefore, $\angle OPQ = 90^\circ$.

$\triangle OPQ$ is a right-angled triangle at P.

By applying the Pythagorean theorem in $\triangle OPQ$, we have:

$OP^2 + PQ^2 = OQ^2$

Substitute the given values:

$OP^2 + (20)^2 = (29)^2$

$OP^2 + 400 = 841$

Now, we need to find $OP^2$. Subtract 400 from both sides:

$OP^2 = 841 - 400$

Performing the subtraction:

$\begin{array}{cc} & 8 & 4 & 1 \\ - & 4 & 0 & 0 \\ \hline & 4 & 4 & 1 \\ \hline \end{array}$

So, $OP^2 = 441$.

To find OP, take the square root of 441:

$OP = \sqrt{441}$

$OP = 21$

The radius of the circle is $21 \text{ cm}$.

---

Answer:

The radius of the circle is $21 \text{ cm}$.

Given:

Let C be a circle with center O.

Let AB be a diameter of the circle.

Let $l$ be the tangent to the circle at point A.

Let $m$ be the tangent to the circle at point B.

---

To Prove:

The tangent $l$ is parallel to the tangent $m$, i.e., $l \parallel m$.

---

Proof:

We know that the radius through the point of contact is perpendicular to the tangent at that point.

For the tangent $l$ at point A, OA is the radius through the point of contact A.

Therefore, OA is perpendicular to the tangent $l$.

where L is a point on the tangent $l$.

Similarly, for the tangent $m$ at point B, OB is the radius through the point of contact B.

Therefore, OB is perpendicular to the tangent $m$.

where M is a point on the tangent $m$.

Now, consider the line segment AB, which is the diameter.

The lines $l$ and $m$ are cut by the transversal line segment AB.

The angles $\angle OAL$ and $\angle OBM$ are alternate interior angles formed by the transversal AB intersecting lines $l$ and $m$.

We have $\angle OAL = 90^\circ$ and $\angle OBM = 90^\circ$.

Thus, $\angle OAL = \angle OBM$.

Since the alternate interior angles formed by the transversal AB with lines $l$ and $m$ are equal, the lines $l$ and $m$ must be parallel.

Therefore, the tangents drawn at the ends of a diameter of a circle are parallel to each other.

Hence Proved.

Given:

PA and PB are tangents from point P to a circle with center O.

The angle between the tangents is $70^\circ$, so $\angle APB = 70^\circ$.

---

To Find:

The measure of $\angle POA$.

---

Solution:

We know that the line joining the center to the external point (OP) bisects the angle between the two tangents from the point to the circle.

Therefore, OP bisects $\angle APB$.

$\angle APO = \angle BPO = \frac{1}{2} \angle APB$

Substituting the given value for $\angle APB$:

$\angle APO = \frac{1}{2} \times 70^\circ = 35^\circ$

We also know that the radius through the point of contact is perpendicular to the tangent at that point.

Therefore, the radius OA is perpendicular to the tangent PA at point A.

$\angle OAP = 90^\circ$

Now, consider the triangle $\triangle POA$. It is a right-angled triangle at A.

The sum of angles in any triangle is $180^\circ$.

In $\triangle POA$, we have:

$\angle POA + \angle OAP + \angle APO = 180^\circ$

Substitute the known angles $\angle OAP = 90^\circ$ and $\angle APO = 35^\circ$:

$\angle POA + 90^\circ + 35^\circ = 180^\circ$

Combine the known angles:

$\angle POA + 125^\circ = 180^\circ$

Subtract $125^\circ$ from both sides to find $\angle POA$:

$\angle POA = 180^\circ - 125^\circ$

$\begin{array}{cc} & 180 \\ - & 125 \\ \hline & \phantom{0}55 \\ \hline \end{array}$

$\angle POA = 55^\circ$

---

Answer:

The measure of $\angle POA$ is $55^\circ$.

Given:

A quadrilateral ABCD circumscribes a circle.

Length of side AB = $8 \text{ cm}$.

Length of side BC = $10 \text{ cm}$.

Length of side CD = $5 \text{ cm}$.

---

To Find:

The length of side AD.

---

Solution:

When a quadrilateral circumscribes a circle, a key property is that the sums of opposite sides are equal.

Let the circle touch the sides AB, BC, CD, and DA at points P, Q, R, and S respectively.

According to the property of tangents drawn from an external point to a circle, the lengths of tangents from an external point to the circle are equal.

Thus, we have:

AP = AS

BP = BQ

CQ = CR

DR = DS

The sides of the quadrilateral can be expressed as:

AB = AP + PB

BC = BQ + QC

CD = CR + RD

DA = DS + SA

Adding opposite sides, we get:

AB + CD = (AP + PB) + (CR + RD)

BC + DA = (BQ + QC) + (DS + SA)

Using the equalities of the tangent segments (AP=AS, PB=BQ, CR=CQ, RD=DS), we can rewrite the first sum:

AB + CD = (AS + BQ) + (CQ + DS)

Rearranging the terms:

AB + CD = (AS + DS) + (BQ + CQ)

Since AS + DS = AD and BQ + CQ = BC, we get:

$AB + CD = AD + BC$

This confirms the property that the sum of opposite sides of a circumscribing quadrilateral is equal.

Now, substitute the given values into this equation:

$8 \text{ cm} + 5 \text{ cm} = AD + 10 \text{ cm}$

$13 \text{ cm} = AD + 10 \text{ cm}$

To find AD, subtract $10 \text{ cm}$ from both sides:

$AD = 13 \text{ cm} - 10 \text{ cm}$

$AD = 3 \text{ cm}$

---

Answer:

The length of side AD is $3 \text{ cm}$.

Given:

Two concentric circles with center O.

Radius of the larger circle, $R = 6 \text{ cm}$.

Radius of the smaller circle, $r = 4 \text{ cm}$.

---

To Find:

The length of the chord of the larger circle that touches the smaller circle.

---

Solution:

Let AB be the chord of the larger circle that touches the smaller circle at point P.

Since AB is tangent to the smaller circle at P, the radius OP of the smaller circle is perpendicular to the tangent AB at P.

Thus, $\angle OPA = \angle OPB = 90^\circ$.

The distance from the center O to the point of tangency P on the smaller circle is the radius of the smaller circle.

$OP = r = 4 \text{ cm}$.

OA is the radius of the larger circle.

$OA = R = 6 \text{ cm}$.

Consider the right-angled triangle $\triangle OPA$ (or $\triangle OPB$).

By the Pythagorean theorem in $\triangle OPA$, we have:

$OP^2 + AP^2 = OA^2$

Substitute the known values:

$(4)^2 + AP^2 = (6)^2$

$16 + AP^2 = 36$

Subtract 16 from both sides to find $AP^2$:

$AP^2 = 36 - 16$

$AP^2 = 20$

Take the square root of both sides to find AP:

$AP = \sqrt{20}$

$AP = \sqrt{4 \times 5}$

$AP = 2\sqrt{5} \text{ cm}$.

We know that a perpendicular from the center to a chord bisects the chord.

Since OP is perpendicular to the chord AB, P is the midpoint of AB.

Therefore, $AB = 2 \times AP$.

Substitute the value of AP:

$AB = 2 \times (2\sqrt{5})$

$AB = 4\sqrt{5} \text{ cm}$.

---

Answer:

The length of the chord of the larger circle which touches the smaller circle is $4\sqrt{5} \text{ cm}$.

Given:

Two concentric circles with center O.

Radius of the larger circle = R.

Radius of the smaller circle = r.

---

To Find:

The length of the chord of the larger circle which touches the smaller circle, in terms of R and r.

---

Solution:

Let AB be the chord of the larger circle that touches the smaller circle at point P.

Since AB is tangent to the smaller circle at P, the radius OP of the smaller circle is perpendicular to the tangent AB at P.

Thus, $\angle OPA = 90^\circ$.

The distance from the center O to the point of tangency P on the smaller circle is the radius of the smaller circle.

$OP = r$.

OA is the radius of the larger circle (connecting the center to a point on the larger circle).

$OA = R$.

Consider the right-angled triangle $\triangle OPA$ at P.

By the Pythagorean theorem in $\triangle OPA$, we have:

$OP^2 + AP^2 = OA^2$

Substitute the values in terms of R and r:

$r^2 + AP^2 = R^2$

Subtract $r^2$ from both sides to find $AP^2$:

$AP^2 = R^2 - r^2$

Take the square root of both sides to find AP:

$AP = \sqrt{R^2 - r^2}$.

We know that a perpendicular from the center to a chord bisects the chord.

Since OP is perpendicular to the chord AB, P is the midpoint of AB.

Therefore, the length of the chord AB is twice the length of AP.

$AB = 2 \times AP$

Substitute the expression for AP:

$AB = 2 \times \sqrt{R^2 - r^2}$

$AB = 2\sqrt{R^2 - r^2}$.

---

Answer:

The length of the chord of the larger circle which touches the smaller circle is $2\sqrt{R^2 - r^2}$.

Given:

Distance of point P from the center O = $13 \text{ cm}$ (i.e., $OP = 13 \text{ cm}$).

Length of the tangent from P to the circle = $12 \text{ cm}$. Let the point of tangency be T, so $PT = 12 \text{ cm}$.

---

To Find:

The diameter of the circle.

---

Solution:

Let O be the center of the circle and let T be the point where the tangent from P touches the circle.

We know that the radius through the point of contact is perpendicular to the tangent at that point.

Therefore, the radius OT is perpendicular to the tangent PT.

$\angle OTP = 90^\circ$.

Thus, $\triangle OTP$ is a right-angled triangle at T.

The hypotenuse of this triangle is the line segment OP, which connects the external point P to the center O.

By applying the Pythagorean theorem in $\triangle OTP$, we have:

$OT^2 + PT^2 = OP^2$

Substitute the given values:

$OT^2 + (12)^2 = (13)^2$

$OT^2 + 144 = 169$

Now, we need to find $OT^2$. Subtract 144 from both sides:

$OT^2 = 169 - 144$

Performing the subtraction:

$\begin{array}{cc} & 1 & 6 & 9 \\ - & 1 & 4 & 4 \\ \hline & \phantom{0}2 & 5 \\ \hline \end{array}$

So, $OT^2 = 25$.

Take the square root of 25 to find the radius OT:

$OT = \sqrt{25}$

$OT = 5 \text{ cm}$.

The radius of the circle is $5 \text{ cm}$.

The diameter of the circle is twice the radius.

Diameter = $2 \times \text{Radius}$

Diameter = $2 \times 5 \text{ cm}$

Diameter = $10 \text{ cm}$.

---

Answer:

The diameter of the circle is $10 \text{ cm}$.

Given:

Radius of the circle, $OA = OB = 5 \text{ cm}$.

Angle between the tangents PA and PB from an external point P, $\angle APB = 60^\circ$.

---

To Find:

The length of the tangent from P to the circle (either PA or PB).

---

Solution:

Let O be the center of the circle and PA and PB be the tangents from point P to the circle, with A and B as the points of contact.

We know that the line segment connecting the external point P to the center O (OP) bisects the angle between the two tangents.

Therefore, OP bisects $\angle APB$.

$\angle APO = \angle BPO = \frac{1}{2} \angle APB$

Substituting the given value of $\angle APB = 60^\circ$:

$\angle APO = \frac{1}{2} \times 60^\circ = 30^\circ$

We also know that the radius drawn to the point of contact is perpendicular to the tangent at that point.

So, the radius OA is perpendicular to the tangent PA.

$\angle OAP = 90^\circ$

Now, consider the right-angled triangle $\triangle OAP$. It is right-angled at A.

We have the radius $OA = 5 \text{ cm}$ and the angle $\angle APO = 30^\circ$. We need to find the length of the tangent PA, which is the adjacent side to the angle $\angle APO$ in $\triangle OAP$. OA is the opposite side.

We can use the trigonometric ratio $\tan$:

$\tan(\angle APO) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{OA}{PA}$

Substitute the values:

$\tan(30^\circ) = \frac{5}{PA}$

We know that the value of $\tan(30^\circ)$ is $\frac{1}{\sqrt{3}}$.

$\frac{1}{\sqrt{3}} = \frac{5}{PA}$

To find PA, cross-multiply:

$PA \times 1 = 5 \times \sqrt{3}$

$PA = 5\sqrt{3}$

The length of the tangent from P to the circle is $5\sqrt{3} \text{ cm}$.

---

Answer:

The length of the tangent from P to the circle is $5\sqrt{3} \text{ cm}$.

Given:

PA and PB are tangents from an external point P to a circle with center O.

The angle between the tangents is $60^\circ$, so $\angle APB = 60^\circ$.

---

To Find:

The measure of $\angle AOB$.

---

Solution:

We know that the radius through the point of contact is perpendicular to the tangent at that point.

Therefore, the radius OA is perpendicular to the tangent PA at A, and the radius OB is perpendicular to the tangent PB at B.

So, we have:

$\angle OAP = 90^\circ$

$\angle OBP = 90^\circ$

Now, consider the quadrilateral PAOB.

The sum of interior angles in a quadrilateral is $360^\circ$.

In quadrilateral PAOB, the sum of angles is:

$\angle APB + \angle PAO + \angle AOB + \angle OBP = 360^\circ$

Substitute the known values:

$60^\circ + 90^\circ + \angle AOB + 90^\circ = 360^\circ$

Combine the known angles:

$60^\circ + 180^\circ + \angle AOB = 360^\circ$

$240^\circ + \angle AOB = 360^\circ$

Subtract $240^\circ$ from both sides to find $\angle AOB$:

$\angle AOB = 360^\circ - 240^\circ$

$\begin{array}{cc} & 3 & 6 & 0 \\ - & 2 & 4 & 0 \\ \hline & 1 & 2 & 0 \\ \hline \end{array}$

$\angle AOB = 120^\circ$.

---

Alternate Solution:

We know that the line segment connecting the external point P to the center O (OP) bisects the angle between the two tangents and also the angle subtended by the points of contact at the center.

Therefore, OP bisects $\angle APB$ and $\angle AOB$.

So, $\angle APO = \angle BPO = \frac{1}{2} \angle APB$ and $\angle AOP = \angle BOP = \frac{1}{2} \angle AOB$.

Given $\angle APB = 60^\circ$, we have:

$\angle APO = \frac{1}{2} \times 60^\circ = 30^\circ$.

In $\triangle OAP$, OA is the radius and PA is the tangent at A. Thus, $\angle OAP = 90^\circ$ (Radius is perpendicular to the tangent at the point of contact).

Now consider the sum of angles in $\triangle OAP$:

$\angle AOP + \angle OAP + \angle APO = 180^\circ$

Substitute the known values:

$\angle AOP + 90^\circ + 30^\circ = 180^\circ$

$\angle AOP + 120^\circ = 180^\circ$

$\angle AOP = 180^\circ - 120^\circ = 60^\circ$.

Since $\angle AOB = 2 \times \angle AOP$, we have:

$\angle AOB = 2 \times 60^\circ = 120^\circ$.

---

Answer:

The measure of $\angle AOB$ is $120^\circ$.

Given:

A quadrilateral ABCD circumscribes a circle.

The sum of the lengths of two opposite sides is $AB + CD = 13 \text{ cm}$.

---

To Find:

The perimeter of the quadrilateral ABCD.

---

Solution:

When a quadrilateral circumscribes a circle, the sums of opposite sides are equal.

This means that the sum of the lengths of one pair of opposite sides is equal to the sum of the lengths of the other pair of opposite sides.

So, for quadrilateral ABCD circumscribing a circle, we have:

$AB + CD = AD + BC$

We are given that $AB + CD = 13 \text{ cm}$.

Therefore, it must also be true that:

$AD + BC = 13 \text{ cm}$.

The perimeter of a quadrilateral is the sum of the lengths of all its four sides.

Perimeter of ABCD = $AB + BC + CD + AD$

We can group the terms: Perimeter = $(AB + CD) + (BC + AD)$.

Substitute the known values for the sums of opposite sides:

Perimeter = $(13 \text{ cm}) + (13 \text{ cm})$

Perimeter = $26 \text{ cm}$.

---

Answer:

The perimeter of the quadrilateral ABCD is $26 \text{ cm}$.

The relationship between the tangent at a point on a circle and the radius through the point of contact is that they are perpendicular to each other.

---

This means that the angle formed between the tangent line and the radius at the point where the tangent touches the circle is always $90^\circ$.

Given:

Two parallel tangents to a circle are $12 \text{ cm}$ apart.

Let the distance between the parallel tangents be $d = 12 \text{ cm}$.

---

To Find:

The radius of the circle.

---

Solution:

Consider a circle with center O.

Let $l_1$ and $l_2$ be two parallel tangents to the circle.

A tangent touches the circle at exactly one point. Let the points of contact for $l_1$ and $l_2$ be A and B, respectively.

Since the tangents are parallel, the points of contact A and B must lie at the opposite ends of a diameter of the circle.

The radius OA is perpendicular to the tangent $l_1$ at A.

The radius OB is perpendicular to the tangent $l_2$ at B.

The line segment AB passes through the center O and connects the two points of contact. AB is the diameter of the circle.

The distance between two parallel lines is the length of a perpendicular segment connecting them.

The line segment AB is perpendicular to both parallel tangents $l_1$ and $l_2$.

Therefore, the distance between the two parallel tangents is equal to the length of the diameter AB.

Distance between tangents = Diameter of the circle

$12 \text{ cm} = \text{Diameter}$

The diameter of the circle is $12 \text{ cm}$.

The radius of a circle is half of its diameter.

Radius $= \frac{\text{Diameter}}{2}$

Radius $= \frac{12 \text{ cm}}{2}$

Radius $= 6 \text{ cm}$.

---

Answer:

The radius of the circle is $6 \text{ cm}$.

Given:

Tangents are drawn from an external point P to a circle.

The length of these tangents is $8 \text{ cm}$.

---

To Find:

The length of a tangent from the same point P to the same circle along a different direction.

---

Solution:

According to a fundamental theorem in geometry, the lengths of tangents drawn from an external point to a circle are always equal.

Let P be the external point and let PA and PB be two different tangents drawn from P to the circle, with A and B being the points of contact on the circle.

According to the theorem:

PA = PB

The question states that the length of tangents from point P to the circle is $8 \text{ cm}$. This means that any tangent segment drawn from P to the circle will have a length of $8 \text{ cm}$.

So, if we draw another tangent from P to the same circle (assuming it's a distinct tangent, which implies a different point of contact), its length will also be equal to the length of the given tangent.

Therefore, the length of the tangent from P to the same circle along a different direction is also $8 \text{ cm}$.

---

Answer:

The length of the tangent from P to the same circle along a different direction is $8 \text{ cm}$.

Given:

AP and AQ are tangents from an external point A to a circle with center O.

The angle between the tangents is $\angle PAQ = 50^\circ$.

---

To Find:

The measure of $\angle POQ$.

---

Solution:

P and Q are the points of contact of the tangents with the circle.

We know that the radius drawn to the point of contact is perpendicular to the tangent at that point.

Therefore, the radius OP is perpendicular to the tangent AP at P, and the radius OQ is perpendicular to the tangent AQ at Q.

So, we have:

$\angle OPA = 90^\circ$

$\angle OQA = 90^\circ$

Now, consider the quadrilateral APAOQ.

The sum of interior angles in a quadrilateral is $360^\circ$.

In quadrilateral APAOQ, the sum of angles is:

$\angle PAQ + \angle OPA + \angle POQ + \angle OQA = 360^\circ$

Substitute the known values:

$50^\circ + 90^\circ + \angle POQ + 90^\circ = 360^\circ$

Combine the known angles:

$50^\circ + 180^\circ + \angle POQ = 360^\circ$

$230^\circ + \angle POQ = 360^\circ$

Subtract $230^\circ$ from both sides to find $\angle POQ$:

$\angle POQ = 360^\circ - 230^\circ$

$\begin{array}{cc} & 3 & 6 & 0 \\ - & 2 & 3 & 0 \\ \hline & 1 & 3 & 0 \\ \hline \end{array}$

$\angle POQ = 130^\circ$.

---

Alternate Solution:

We know that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the center.

That is, $\angle PAQ + \angle POQ = 180^\circ$.

We are given $\angle PAQ = 50^\circ$.

Substitute this value into the equation:

$50^\circ + \angle POQ = 180^\circ$

Subtract $50^\circ$ from both sides:

$\angle POQ = 180^\circ - 50^\circ$

$\angle POQ = 130^\circ$.

---

Answer:

The measure of $\angle POQ$ is $130^\circ$.

Given:

A triangle ABC has a circle inscribed in it, touching sides AB, BC, and CA at points D, E, and F respectively.

Length of side AB = $10 \text{ cm}$.

Length of side BC = $12 \text{ cm}$.

Length of side AC = $14 \text{ cm}$.

Length of segment AD = x.

---

To Find:

The value of x.

---

Solution:

We know that the lengths of tangents drawn from an external point to a circle are equal.

Considering the vertices of the triangle as external points, we have the following pairs of equal tangents:

From vertex A: AD and AF are tangents to the circle.

So, $AD = AF = x$.

From vertex B: BD and BE are tangents to the circle.

So, $BD = BE$.

From vertex C: CE and CF are tangents to the circle.

So, $CE = CF$.

We can express the lengths of the sides of the triangle in terms of these tangent segments:

$AB = AD + DB$

$BC = BE + EC$

$AC = AF + FC$

Substitute the given lengths and the equalities of the tangent segments:

$AB = x + BD \implies 10 = x + BD \implies BD = 10 - x$

Since $BD = BE$, we have $BE = 10 - x$.

$AC = x + CF \implies 14 = x + CF \implies CF = 14 - x$

Since $CF = CE$, we have $CE = 14 - x$.

Now consider the side BC:

$BC = BE + CE$

Substitute the expressions for BE and CE:

$12 = (10 - x) + (14 - x)$

$12 = 10 - x + 14 - x$

Combine the constant terms and the x terms:

$12 = (10 + 14) + (-x - x)$

$12 = 24 - 2x$

Now, we need to solve for x. Add 2x to both sides:

$12 + 2x = 24$

Subtract 12 from both sides:

$2x = 24 - 12$

$2x = 12$

Divide both sides by 2:

$x = \frac{12}{2}$

$x = 6$.

The value of x is 6 cm.

We can verify the lengths of the tangent segments:

AD = AF = x = 6 cm.

BD = BE = $10 - x = 10 - 6 = 4$ cm.

CE = CF = $14 - x = 14 - 6 = 8$ cm.

Check the side lengths:

AB = AD + DB = $6 + 4 = 10$ cm (Correct)

BC = BE + EC = $4 + 8 = 12$ cm (Correct)

AC = AF + FC = $6 + 8 = 14$ cm (Correct)

---

Answer:

The value of x is 6.

Given:

TA and TB are tangents from an external point T to a circle with center O.

The angle subtended by the points of contact at the center is $\angle AOB = 110^\circ$.

---

To Find:

The measure of the angle between the tangents, $\angle ATB$.

---

Solution:

A and B are the points of contact of the tangents with the circle.

We know that the radius drawn to the point of contact is perpendicular to the tangent at that point.

Therefore, the radius OA is perpendicular to the tangent TA at A, and the radius OB is perpendicular to the tangent TB at B.

So, we have:

$\angle OAT = 90^\circ$

$\angle OBT = 90^\circ$

Now, consider the quadrilateral TAOB.

The sum of interior angles in a quadrilateral is $360^\circ$.

In quadrilateral TAOB, the sum of angles is:

$\angle ATB + \angle OAT + \angle AOB + \angle OBT = 360^\circ$

Substitute the known values:

$\angle ATB + 90^\circ + 110^\circ + 90^\circ = 360^\circ$

Combine the known angles:

$\angle ATB + 290^\circ = 360^\circ$

Subtract $290^\circ$ from both sides to find $\angle ATB$:

$\angle ATB = 360^\circ - 290^\circ$

$\begin{array}{cc} & 3 & 6 & 0 \\ - & 2 & 9 & 0 \\ \hline & \phantom{0}7 & 0 \\ \hline \end{array}$

$\angle ATB = 70^\circ$.

---

Alternate Solution:

We know that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the center.

That is, $\angle ATB + \angle AOB = 180^\circ$.

We are given $\angle AOB = 110^\circ$.

Substitute this value into the equation:

$\angle ATB + 110^\circ = 180^\circ$

Subtract $110^\circ$ from both sides:

$\angle ATB = 180^\circ - 110^\circ$

$\angle ATB = 70^\circ$.

---

Answer:

The measure of $\angle ATB$ is $70^\circ$.

Given:

Radius of the circle, $OT = 10 \text{ cm}$.

Distance of the external point P from the center O, $OP = 26 \text{ cm}$.

---

To Find:

The length of the tangent from P to the circle, $PT$.

---

Solution:

Let O be the center of the circle and T be the point of contact of the tangent drawn from the external point P.

We know that the radius through the point of contact is perpendicular to the tangent at that point.

Therefore, the radius OT is perpendicular to the tangent PT.

$\angle OTP = 90^\circ$.

Thus, $\triangle OTP$ is a right-angled triangle at T.

The hypotenuse of this triangle is the line segment OP, which connects the external point P to the center O.

By applying the Pythagorean theorem in $\triangle OTP$, we have:

$OT^2 + PT^2 = OP^2$

Substitute the given values:

$(10)^2 + PT^2 = (26)^2$

$100 + PT^2 = 676$

Now, we need to find $PT^2$. Subtract 100 from both sides:

$PT^2 = 676 - 100$

$PT^2 = 576$

Take the square root of 576 to find the length of the tangent PT:

$PT = \sqrt{576}$

$PT = 24$

Since the given lengths are in cm, the length of the tangent PT is $24 \text{ cm}$.

---

Answer:

The length of the tangent from P to the circle is $24 \text{ cm}$.

Given:

PA and PB are tangents drawn from an external point P to a circle with center O.

A and B are the points of contact.

---

To Prove:

The points A, O, B, and P are concyclic.

---

Proof:

We know that the radius through the point of contact is perpendicular to the tangent at that point.

For the tangent PA at point A, OA is the radius through A.

Therefore, OA is perpendicular to PA.

$\angle OAP = 90^\circ$

(Radius is perpendicular to tangent at point of contact)

Similarly, for the tangent PB at point B, OB is the radius through B.

Therefore, OB is perpendicular to PB.

$\angle OBP = 90^\circ$

(Radius is perpendicular to tangent at point of contact)

Now, consider the quadrilateral APBO. The four vertices are A, P, B, and O.

The angles at vertices A and B are $\angle OAP$ and $\angle OBP$. These are opposite angles in the quadrilateral APBO.

Let's find the sum of these opposite angles:

$\angle OAP + \angle OBP = 90^\circ + 90^\circ$

$\angle OAP + \angle OBP = 180^\circ$

We know that if the sum of a pair of opposite angles of a quadrilateral is $180^\circ$, then the quadrilateral is cyclic.

Since the sum of opposite angles $\angle OAP$ and $\angle OBP$ in quadrilateral APBO is $180^\circ$, the quadrilateral APBO is a cyclic quadrilateral.

This means that all four vertices A, P, B, and O lie on the same circle.

Therefore, the points A, O, B, and P are concyclic.

Hence Proved.

Given:

Radius of the circle = $r$.

Distance of the external point P from the center O = $d$ (i.e., $OP = d$).

---

To Find:

The length of the tangent from P to the circle.

---

Solution:

Let O be the center of the circle and T be the point of contact of the tangent drawn from the external point P.

The length of the tangent from P to the circle is the length of the line segment PT.

We know that the radius through the point of contact is perpendicular to the tangent at that point.

Therefore, the radius OT is perpendicular to the tangent PT.

$\angle OTP = 90^\circ$.

Thus, $\triangle OTP$ is a right-angled triangle at T.

The lengths of the sides of the triangle are:

Hypotenuse = Distance from O to P = $OP = d$.

One leg = Radius of the circle = $OT = r$.

The other leg = Length of the tangent = $PT$.

By applying the Pythagorean theorem in $\triangle OTP$, we have:

$OT^2 + PT^2 = OP^2$

Substitute the given values in terms of $d$ and $r$:

$r^2 + PT^2 = d^2$

Now, we need to find $PT^2$. Subtract $r^2$ from both sides:

$PT^2 = d^2 - r^2$

Take the square root of both sides to find the length of the tangent PT:

$PT = \sqrt{d^2 - r^2}$.

Note that for the tangent to exist from an external point, the distance $d$ must be greater than the radius $r$, so $d > r$, which ensures $d^2 - r^2$ is positive.

---

Answer:

The length of the tangent from P to the circle is $\sqrt{d^2 - r^2}$.

From a point inside a circle, zero tangents can be drawn to the circle.

---

Justification:

A tangent to a circle is defined as a straight line that touches the circle at exactly one point.

Let O be the center of the circle and $r$ be its radius.

Consider a point P located inside the circle. This means the distance of the point P from the center O is less than the radius, i.e., $OP < r$.

Any straight line passing through a point P that is inside the circle will always intersect the circle at two distinct points.

This is because the perpendicular distance from the center O to any line passing through P must be less than or equal to OP. Since $OP < r$, the perpendicular distance from O to any such line will be less than $r$. A line with a perpendicular distance less than the radius from the center is a secant, not a tangent.

Since no line passing through a point inside the circle can intersect the circle at just one point, no tangent can be drawn from a point inside the circle.

Given:

A circle with center O.

An external point P outside the circle.

Two tangents PA and PB are drawn from P to the circle, touching the circle at points A and B respectively.

---

To Prove:

The lengths of the tangents PA and PB are equal, i.e., $PA = PB$.

---

Construction Required:

Join the center O to the points of contact A and B (OA and OB).

Join the center O to the external point P (OP).

---

Proof:

We know that the radius through the point of contact is perpendicular to the tangent at that point.

For the tangent PA at point A, OA is the radius through A.

Similarly, for the tangent PB at point B, OB is the radius through B.

Now, consider the two triangles $\triangle OAP$ and $\triangle OBP$.

In these two triangles, we have:

1. $OA = OB$

2. OP = OP

3. $\angle OAP = \angle OBP$

Since $\triangle OAP$ and $\triangle OBP$ are both right-angled triangles, with the hypotenuse OP being common and one side (OA = OB) being equal, the triangles are congruent by the RHS (Right angle - Hypotenuse - Side) congruence criterion.

Therefore, $\triangle OAP \cong \triangle OBP$.

By CPCTC (Corresponding Parts of Congruent Triangles), the corresponding sides of the congruent triangles are equal.

Thus, we have:

$PA = PB$

This proves that the lengths of the tangents drawn from the external point P to the circle are equal.

Hence Proved.

Given:

A circle with center O.

An external point P outside the circle.

Tangents PA and PB are drawn from P to the circle, touching the circle at points A and B respectively.

---

To Prove:

The angle between the tangents ($\angle APB$) and the angle subtended by the chord of contact AB at the center ($\angle AOB$) are supplementary.

That is, $\angle APB + \angle AOB = 180^\circ$.

---

Proof:

We know that the radius through the point of contact is perpendicular to the tangent at that point.

For the tangent PA at point A, OA is the radius through A.

Similarly, for the tangent PB at point B, OB is the radius through B.

Now, consider the quadrilateral PAOB.

The sum of interior angles in any quadrilateral is $360^\circ$.

The angles in the quadrilateral PAOB are $\angle APB$, $\angle OAP$, $\angle AOB$, and $\angle OBP$.

Therefore, the sum of these angles is:

$\angle APB + \angle OAP + \angle AOB + \angle OBP = 360^\circ$

Substitute the values of the right angles $\angle OAP$ and $\angle OBP$:

$\angle APB + 90^\circ + \angle AOB + 90^\circ = 360^\circ$

Combine the constant terms:

$\angle APB + \angle AOB + 180^\circ = 360^\circ$

Subtract $180^\circ$ from both sides of the equation:

$\angle APB + \angle AOB = 360^\circ - 180^\circ$

$\angle APB + \angle AOB = 180^\circ$

Since the sum of $\angle APB$ and $\angle AOB$ is $180^\circ$, these two angles are supplementary.

This proves that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the center.

Hence Proved.

Given:

A triangle ABC circumscribes a circle with center O and radius $r = 4 \text{ cm}$.

The circle touches side BC at point D.

Length of segment BD = $8 \text{ cm}$.

Length of segment DC = $6 \text{ cm}$.

---

To Find:

The lengths of sides AB and AC.

---

Solution:

Let the circle touch the sides AB and AC at points E and F respectively.

We know that the lengths of tangents drawn from an external point to a circle are equal.

From vertex B, BD and BE are tangents to the circle.

$BE = BD$

From vertex C, CD and CF are tangents to the circle.

$CF = CD$

From vertex A, AE and AF are tangents to the circle.

$AE = AF$

Let $AE = AF = x \text{ cm}$.

Now, we can express the lengths of the sides of the triangle:

$AB = AE + EB = x + 8$ cm

$BC = BD + DC = 8 + 6 = 14$ cm

$AC = AF + FC = x + 6$ cm

---

The semi-perimeter $s$ of $\triangle ABC$ is:

$s = \frac{AB + BC + AC}{2}$

$s = \frac{(x+8) + 14 + (x+6)}{2}$

$s = \frac{2x + 28}{2}$

$s = x + 14$ cm.

---

The area of $\triangle ABC$ can be calculated using two methods:

Method 1: Using Heron's formula

Area $= \sqrt{s(s-a)(s-b)(s-c)}$, where $a=BC, b=AC, c=AB$.

$s-a = (x+14) - 14 = x$

$s-b = (x+14) - (x+6) = x+14-x-6 = 8$

$s-c = (x+14) - (x+8) = x+14-x-8 = 6$

Area $= \sqrt{(x+14)(x)(8)(6)}$

Area $= \sqrt{48x(x+14)}$

---

Method 2: Using the formula Area = $rs$

Where $r$ is the inradius and $s$ is the semi-perimeter.

We are given $r=4$ cm and we found $s = x+14$.

Area $= 4(x+14)$.

---

Equating the two expressions for the area:

$\sqrt{48x(x+14)} = 4(x+14)$

Square both sides:

$48x(x+14) = 16(x+14)^2$

Divide both sides by $16(x+14)$ (since $x$ must be positive, $x+14 \neq 0$):

$\frac{\cancel{48}^{3}x\cancel{(x+14)}}{\cancel{16}_{1}\cancel{(x+14)}} = \frac{\cancel{16}(x+14)^{\cancel{2}}}{\cancel{16}(x+14)}$

$3x = x+14$

Subtract x from both sides:

$3x - x = 14$

$2x = 14$

Divide by 2:

$x = \frac{14}{2}$

$x = 7$

The value of x is 7 cm.

---

Now we can find the lengths of sides AB and AC:

$AB = x + 8 = 7 + 8 = 15$ cm.

$AC = x + 6 = 7 + 6 = 13$ cm.

---

Answer:

The length of side AB is $15 \text{ cm}$.

The length of side AC is $13 \text{ cm}$.

Given:

ABCD is a parallelogram circumscribing a circle.

Let the circle touch the sides AB, BC, CD, and DA at points P, Q, R, and S respectively.

---

To Prove:

Parallelogram ABCD is a rhombus.

---

Proof:

Since ABCD is a parallelogram, we know that its opposite sides are equal in length.

$AB = CD$

$BC = DA$

When a quadrilateral circumscribes a circle, the sums of opposite sides are equal.

Therefore, for the quadrilateral ABCD circumscribing the circle, we have:

$AB + CD = BC + DA$

Substitute the properties of the parallelogram into this equation.

$AB + AB = BC + DA$

$2AB = BC + DA$

Similarly, since $BC = DA$, we can replace DA with BC on the right side:

$2AB = BC + BC$

$2AB = 2BC$

Divide both sides by 2:

$AB = BC$

So, we have shown that two adjacent sides of the parallelogram (AB and BC) are equal.

In a parallelogram, if a pair of adjacent sides are equal, then all four sides are equal.

Since $AB = BC$ and opposite sides are equal ($AB = CD$ and $BC = DA$), it follows that:

$AB = BC = CD = DA$

A parallelogram with all four sides of equal length is defined as a rhombus.

Therefore, a parallelogram circumscribing a circle is a rhombus.

Hence Proved.

Given:

A circle with center O.

An external point P outside the circle.

Tangents PA and PB are drawn from P to the circle, touching the circle at points A and B respectively.

---

To Prove:

The line segment OP is the perpendicular bisector of the line segment AB.

---

Construction Required:

Join OA, OB, and OP.

Let the line segment OP intersect the line segment AB at point M.

---

Proof:

We know that the radius through the point of contact is perpendicular to the tangent at that point.

For the tangent PA at point A, OA is the radius through A.

Similarly, for the tangent PB at point B, OB is the radius through B.

Consider the two triangles $\triangle OAP$ and $\triangle OBP$.

In these two triangles, we have:

1. $OA = OB$

2. OP = OP

3. $\angle OAP = \angle OBP$

Since $\triangle OAP$ and $\triangle OBP$ are both right-angled triangles, with the hypotenuse OP being common and one side (OA = OB) being equal, the triangles are congruent by the RHS (Right angle - Hypotenuse - Side) congruence criterion.

Therefore, $\triangle OAP \cong \triangle OBP$.

By CPCTC (Corresponding Parts of Congruent Triangles), the corresponding parts of congruent triangles are equal.

From the congruence of $\triangle OAP$ and $\triangle OBP$, we get:

$PA = PB$

And $\angle APO = \angle BPO$. This means OP bisects $\angle APB$.

---

Now, consider the two triangles $\triangle APM$ and $\triangle BPM$.

In these two triangles, we have:

1. $PA = PB$

2. $\angle APM = \angle BPM$

3. PM = PM

Therefore, $\triangle APM \cong \triangle BPM$ by the SAS (Side - Angle - Side) congruence criterion.

By CPCTC (Corresponding Parts of Congruent Triangles), the corresponding parts of congruent triangles are equal.

From the congruence of $\triangle APM$ and $\triangle BPM$, we get:

$AM = BM$

This implies that point M is the midpoint of the line segment AB. Since M lies on OP, the line segment OP bisects the line segment AB.

We also get from the congruence:

$\angle AMP = \angle BMP$

Angles $\angle AMP$ and $\angle BMP$ form a linear pair on the straight line segment AB.

The sum of angles in a linear pair is $180^\circ$.

So, $\angle AMP + \angle BMP = 180^\circ$.

Since $\angle AMP = \angle BMP$, we can write:

$\angle AMP + \angle AMP = 180^\circ$

$2 \angle AMP = 180^\circ$

$\angle AMP = \frac{180^\circ}{2}$

$\angle AMP = 90^\circ$

Since $\angle AMP = 90^\circ$, the line segment PM is perpendicular to the line segment AB. As M lies on OP, this means the line segment OP is perpendicular to the line segment AB.

---

We have shown that OP bisects AB (AM = BM) and OP is perpendicular to AB ($\angle AMP = 90^\circ$).

Therefore, OP is the perpendicular bisector of the line segment AB.

Hence Proved.

Given:

A right-angled triangle ABC, with the right angle at B ($\angle B = 90^\circ$).

A circle with radius $r$ and center O is inscribed in $\triangle ABC$, touching the sides.

Lengths of the sides are AB = a, BC = b, and AC = c.

---

To Prove:

The radius of the inscribed circle is given by the formula $r = \frac{a+b-c}{2}$.

---

Construction Required:

Let the inscribed circle touch the sides AB, BC, and CA at points P, Q, and R respectively.

Join the center O to the points of contact P, Q, and R (OP, OQ, OR).

Join the center O to the vertices A, B, and C (OA, OB, OC).

---

Proof:

We know that the radius through the point of contact is perpendicular to the tangent at that point.

Thus, we have:

$OP \perp AB$

$OQ \perp BC$

$OR \perp CA$

So, $\angle OPA = \angle OPB = 90^\circ$, $\angle OQC = \angle OQB = 90^\circ$, and $\angle ORA = \angle ORC = 90^\circ$.

Also, the lengths of the radii are equal: $OP = OQ = OR = r$.

Consider the quadrilateral OPBQ. We have:

The sum of angles in a quadrilateral is $360^\circ$. In OPBQ:

$\angle POQ + \angle OPB + \angle PBQ + \angle OQB = 360^\circ$

$\angle POQ + 90^\circ + 90^\circ + 90^\circ = 360^\circ$

$\angle POQ + 270^\circ = 360^\circ$

$\angle POQ = 90^\circ$.

Since all angles in quadrilateral OPBQ are $90^\circ$, it is a rectangle. Furthermore, since adjacent sides $OP$ and $OQ$ are radii of the same circle ($OP = OQ = r$), all four sides are equal.

Therefore, OPBQ is a square with side length $r$.

This means $BP = r$ and $BQ = r$.

---

Now, we use the property that the lengths of tangents drawn from an external point to a circle are equal.

From vertex A, AP and AR are tangents to the circle.

$AP = AR$.

From vertex C, CQ and CR are tangents to the circle.

$CQ = CR$.

We know the length of side AB is $a$. $AB = AP + PB$. Since $PB = r$, we have:

$a = AP + r \implies AP = a - r$.

Since $AP = AR$, we get $AR = a - r$.

We know the length of side BC is $b$. $BC = BQ + QC$. Since $BQ = r$, we have:

$b = r + QC \implies QC = b - r$.

Since $QC = CR$, we get $CR = b - r$.

The length of the hypotenuse AC is $c$. AC is composed of the tangent segments AR and RC from vertex A and C respectively.

$AC = AR + RC$

Substitute the expressions for AR and RC in terms of $a$, $b$, and $r$:

$c = (a - r) + (b - r)$

$c = a + b - 2r$

Now, we need to solve this equation for $r$. Add $2r$ to both sides:

$c + 2r = a + b$

Subtract $c$ from both sides:

$2r = a + b - c$

Divide both sides by 2:

$r = \frac{a+b-c}{2}$

This proves the formula for the radius of the inscribed circle in a right-angled triangle.

Hence Proved.

Given:

PQ and PR are tangents from an external point P to a circle with center O.

The angle between the tangents is $\angle QPR = 60^\circ$.

The radius of the circle is $OQ = OR = 6 \text{ cm}$.

---

To Find:

The length of the tangents PQ and PR.

The length of the chord QR.

---

Solution:

We know that the lengths of tangents drawn from an external point to a circle are equal.

Therefore, $PQ = PR$.

We also know that the line segment connecting the external point P to the center O (OP) bisects the angle between the two tangents.

So, OP bisects $\angle QPR$.

$\angle QPO = \angle RPO = \frac{1}{2} \angle QPR$

Substituting the given value $\angle QPR = 60^\circ$:

$\angle QPO = \frac{1}{2} \times 60^\circ = 30^\circ$

We know that the radius drawn to the point of contact is perpendicular to the tangent at that point.

So, the radius OQ is perpendicular to the tangent PQ at Q.

$\angle OQP = 90^\circ$

Now, consider the right-angled triangle $\triangle OQP$. It is right-angled at Q.

We have the radius $OQ = 6 \text{ cm}$ and the angle $\angle QPO = 30^\circ$. We need to find the length of the tangent PQ.

In $\triangle OQP$, OQ is the side opposite to $\angle QPO$, and PQ is the side adjacent to $\angle QPO$. We can use the tangent ratio:

$\tan(\angle QPO) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{OQ}{PQ}$

Substitute the values:

$\tan(30^\circ) = \frac{6}{PQ}$

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$.

$\frac{1}{\sqrt{3}} = \frac{6}{PQ}$

Cross-multiply to find PQ:

$PQ \times 1 = 6 \times \sqrt{3}$

$PQ = 6\sqrt{3}$ cm.

Since $PQ = PR$, the length of both tangents is $6\sqrt{3} \text{ cm}$.

---

Now, let's find the length of the chord QR.

We know that the line joining the center O to the external point P (OP) bisects the angle between the tangents ($\angle QPR$) and also bisects the angle subtended by the chord of contact at the center ($\angle QOR$). OP is also the perpendicular bisector of the chord QR.

In quadrilateral PQOR, the sum of angles is $360^\circ$.

$\angle QPR + \angle OQP + \angle QOR + \angle ORP = 360^\circ$

Substitute the known values ($\angle QPR = 60^\circ$, $\angle OQP = 90^\circ$, $\angle ORP = 90^\circ$):

$60^\circ + 90^\circ + \angle QOR + 90^\circ = 360^\circ$

$240^\circ + \angle QOR = 360^\circ$

$\angle QOR = 360^\circ - 240^\circ = 120^\circ$.

Since OP bisects $\angle QOR$, let M be the point where OP intersects QR.

$\angle QOM = \angle ROM = \frac{1}{2} \angle QOR = \frac{1}{2} \times 120^\circ = 60^\circ$.

Also, OP is perpendicular to QR, so $\triangle OMQ$ is a right-angled triangle at M.

In right $\triangle OMQ$, we have $OQ = 6 \text{ cm}$ and $\angle QOM = 60^\circ$. We need to find QM.

Using the sine ratio in $\triangle OMQ$:

$\sin(\angle QOM) = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{QM}{OQ}$

Substitute the values:

$\sin(60^\circ) = \frac{QM}{6}$

We know that $\sin(60^\circ) = \frac{\sqrt{3}}{2}$.

$\frac{\sqrt{3}}{2} = \frac{QM}{6}$

Solve for QM:

$QM = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}$ cm.

Since M is the midpoint of QR, the length of the chord QR is $2 \times QM$.

$QR = 2 \times (3\sqrt{3}) = 6\sqrt{3}$ cm.

---

Alternate Method for QR:

Consider $\triangle PQR$. We know that $PQ = PR$ (lengths of tangents from an external point are equal), so $\triangle PQR$ is an isosceles triangle.

The angle between the equal sides is $\angle QPR = 60^\circ$.

In an isosceles triangle, the base angles are equal: $\angle PQR = \angle PRQ$.

The sum of angles in $\triangle PQR$ is $180^\circ$.

$\angle QPR + \angle PQR + \angle PRQ = 180^\circ$

$60^\circ + \angle PQR + \angle PQR = 180^\circ$

$60^\circ + 2\angle PQR = 180^\circ$

$2\angle PQR = 180^\circ - 60^\circ = 120^\circ$

$\angle PQR = \frac{120^\circ}{2} = 60^\circ$.

Since all angles in $\triangle PQR$ are $60^\circ$ ($\angle QPR = 60^\circ$, $\angle PQR = 60^\circ$, $\angle PRQ = 60^\circ$), $\triangle PQR$ is an equilateral triangle.

Therefore, all sides of $\triangle PQR$ are equal:

$PQ = PR = QR$

Since we found $PQ = 6\sqrt{3}$ cm, the length of the chord QR is also $6\sqrt{3} \text{ cm}$.

---

Answer:

The length of the tangents PQ and PR is $6\sqrt{3} \text{ cm}$.

The length of the chord QR is $6\sqrt{3} \text{ cm}$.

Given:

Two circles with centers O and O' and radii $r_1$ and $r_2$ respectively.

The circles touch externally at point C.

A common tangent AB touches the circle with center O at A and the circle with center O' at B.

---

To Prove:

$AB^2 = 4 r_1 r_2$.

---

Construction Required:

Draw a line segment OP parallel to AB through O, meeting the line segment O'B produced at P.

Join OA, OB, O'A, O'B.

---

Proof:

Since the circles touch externally at C, the line segment connecting the centers passes through C.

$OO' = OC + O'C = r_1 + r_2$.

We know that the radius through the point of contact is perpendicular to the tangent at that point.

Therefore, the radius OA is perpendicular to the tangent AB at A.

Similarly, the radius O'B is perpendicular to the tangent AB at B.

Now, consider the line segment OP drawn parallel to AB through O, meeting O'B (or its extension) at P.

Since OP is parallel to AB and O'B is perpendicular to AB, O'B is also perpendicular to OP.

Thus, $\angle OPB = 90^\circ$.

Consider the quadrilateral AOPB.

$\angle OAB = 90^\circ$, $\angle O'BA = 90^\circ$. AB is parallel to OP.

This forms a rectangle AOPB, with sides AB and O'P?

Let's reconsider the construction. Draw a line through O parallel to AB, meeting the line O'B at P.

Since OP || AB and AB $\perp$ O'B, it follows that OP $\perp$ O'B. Thus, $\angle OPO' = 90^\circ$.

Consider the quadrilateral AOPB.

$\angle OAP = 90^\circ$ (assuming A is the point of contact for O, O'A is not involved yet).

Let's adjust the construction. Draw a line through O parallel to AB, meeting O'B at P.

A is the point of contact for the circle with center O, so OA $\perp$ AB. $\angle OAB = 90^\circ$. OA = $r_1$.

B is the point of contact for the circle with center O', so O'B $\perp$ AB. $\angle O'BA = 90^\circ$. O'B = $r_2$.

Draw a line through O parallel to AB. Let this line intersect the line segment O'B at P.

Since OP is parallel to AB and AB is perpendicular to O'B, OP is perpendicular to O'B.

Thus, $\angle OPO' = 90^\circ$.

The figure AOPB is a rectangle.

Side AB is parallel to OP. Side OA is parallel to PB?

Let's redraw. Draw a line through O parallel to AB, meeting the line O'B at P.

A O P B forms a rectangle? No.

Draw a line through O parallel to AB. Let it meet O'B at P.

This creates a right-angled triangle $\triangle OO'P$.

Hypotenuse $OO' = r_1 + r_2$.

One leg is $OP$. Since AOPB is a rectangle (OA || PB, AB || OP, $\angle OAP = 90$, $\angle OPB = 90$), OP = AB.

The other leg is $O'P$. The length $O'P$ is the difference between the lengths $O'B$ and PB. Since AOPB is a rectangle, PB = OA = $r_1$.

$O'P = O'B - PB = r_2 - r_1$. (Assuming $r_2 > r_1$ for the diagram, if $r_1 > r_2$ then $O'P = r_1 - r_2$, so $|r_2 - r_1|$)

Apply the Pythagorean theorem in right $\triangle OO'P$:

$OP^2 + O'P^2 = OO'^2$

$AB^2 + (r_2 - r_1)^2 = (r_1 + r_2)^2$

$AB^2 + r_2^2 - 2r_1 r_2 + r_1^2 = r_1^2 + 2r_1 r_2 + r_2^2$

Subtract $r_1^2 + r_2^2$ from both sides:

$AB^2 - 2r_1 r_2 = 2r_1 r_2$

Add $2r_1 r_2$ to both sides:

$AB^2 = 2r_1 r_2 + 2r_1 r_2$

$AB^2 = 4 r_1 r_2$.

This holds regardless of whether $r_1 > r_2$ or $r_2 > r_1$, as $(r_2 - r_1)^2 = (r_1 - r_2)^2$.

Therefore, the length of the common tangent squared is $4 r_1 r_2$.

Hence Proved.

Given:

PA and PB are tangents from external point P to a circle with center O.

A tangent at point M on the circle intersects PA at K and PB at L.

Length of tangent PA = $15 \text{ cm}$.

---

To Find:

The perimeter of triangle PKL.

---

Solution:

We know that the lengths of tangents drawn from an external point to a circle are equal.

Considering point P as the external point and PA and PB as tangents to the circle, we have:

$PA = PB$

Now consider point K on the tangent PA. K is an external point to the circle, and tangents drawn from K are KA and KM (since the tangent at M passes through K).

So, $KA = KM$.

Similarly, consider point L on the tangent PB. L is an external point to the circle, and tangents drawn from L are LB and LM (since the tangent at M passes through L).

So, $LB = LM$.

The perimeter of triangle PKL is the sum of the lengths of its sides:

Perimeter of $\triangle PKL = PK + KL + LP$

The side KL is composed of the tangent segments KM and ML.

$KL = KM + ML$

Substitute this into the perimeter expression:

Perimeter of $\triangle PKL = PK + (KM + ML) + LP$

Now, use the equalities of tangent lengths we found:

Replace KM with KA (since KM = KA):

Perimeter of $\triangle PKL = PK + KA + ML + LP$

Notice that $PK + KA$ is the length of the side PA.

$PK + KA = PA$

Replace ML with LB (since ML = LB):

Perimeter of $\triangle PKL = PA + LB + LP$

Rearrange the terms involving L and P:

Perimeter of $\triangle PKL = PA + (LP + LB)$

Notice that $LP + LB$ is the length of the side PB.

$LP + LB = PB$

Substitute this into the perimeter expression:

Perimeter of $\triangle PKL = PA + PB$

We are given that $PA = 15 \text{ cm}$, and we know that $PB = PA = 15 \text{ cm}$.

Perimeter of $\triangle PKL = 15 \text{ cm} + 15 \text{ cm}$

Perimeter of $\triangle PKL = 30 \text{ cm}$.

---

Answer:

The perimeter of triangle PKL is $30 \text{ cm}$.

Given:

A circle with center O.

A tangent line T'AT touches the circle at point A.

AB is a chord through the point of contact A.

C is a point on the circumference in the alternate segment (the segment that does not contain T').

---

To Prove:

The angle between the tangent and the chord is equal to the angle in the alternate segment.

That is, $\angle TAB = \angle ACB$.

---

Construction Required:

Draw the diameter AD through A.

Join CD and BD.

---

Proof:

There are three cases to consider based on the type of angle $\angle ACB$:

Case 1: The angle $\angle ACB$ is acute.

We know that the radius through the point of contact is perpendicular to the tangent at that point.

Since AD is the diameter, OA is a radius. The tangent T'AT is at A.

The angle subtended by a diameter at any point on the circumference is $90^\circ$.

Consider the right-angled triangle $\triangle ABD$. The sum of angles in a triangle is $180^\circ$.

In $\triangle ABD$:

$\angle DAB + \angle ADB + \angle ABD = 180^\circ$

$\angle DAB + \angle ADB + 90^\circ = 180^\circ$

$\angle DAB + \angle ADB = 90^\circ$ ... (i)

Now, consider the angle $\angle TAD$. It can be written as the sum of $\angle TAB$ and $\angle DAB$.

$\angle TAD = \angle TAB + \angle DAB$

Since $\angle TAD = 90^\circ$, we have:

$\angle TAB + \angle DAB = 90^\circ$ ... (ii)

Comparing equations (i) and (ii):

$\angle DAB + \angle ADB = \angle TAB + \angle DAB$

Subtract $\angle DAB$ from both sides:

$\angle ADB = \angle TAB$

Angles subtended by the same arc AB at the circumference in the same segment are equal.

Points C and D are both on the circumference and are on the same side of the chord AB (in the alternate segment to $\angle TAB$).

From $\angle TAB = \angle ADB$ and $\angle ADB = \angle ACB$, we conclude:

$\angle TAB = \angle ACB$.

This proves the theorem for the acute angle case.

---

Case 2: The angle $\angle ACB$ is a right angle.

If $\angle ACB = 90^\circ$, then the chord AB must be a diameter of the circle.

If AB is a diameter, the tangent at A is perpendicular to the diameter AB.

The angle in the alternate segment is $\angle ACB$.

Therefore, $\angle TAB = \angle ACB = 90^\circ$. The theorem holds.

---

Case 3: The angle $\angle ACB$ is obtuse.

In this case, C is on the minor arc AB. The alternate segment to $\angle TAB$ is the major segment.

Let C be a point on the minor arc. Let D be a point on the major arc such that ACBD forms a cyclic quadrilateral. The angle in the alternate segment corresponding to $\angle TAB$ is $\angle ADB$ (where D is on the major arc). $\angle ADB$ will be acute since $\angle ACB$ is obtuse and $\angle ACB + \angle ADB = 180^\circ$.

From Case 1, we proved that for any point D on the major arc AB, $\angle TAB = \angle ADB$.

Consider the cyclic quadrilateral ACBD. The sum of opposite angles is $180^\circ$.

We also know that the angle $\angle T'AB$ (the other angle between the tangent and the chord) and $\angle TAB$ form a linear pair.

From $\angle TAB = \angle ADB$ (proved in Case 1, holds for D on major arc) and $\angle ACB + \angle ADB = 180^\circ$, we have:

$\angle ACB + \angle TAB = 180^\circ$

Comparing this with $\angle T'AB + \angle TAB = 180^\circ$, we get:

$\angle T'AB = \angle ACB$

So, the angle between the tangent and the chord in one direction ($\angle T'AB$) is equal to the obtuse angle in the segment ($\angle ACB$). The angle between the tangent and the chord in the other direction ($\angle TAB$) is equal to the acute angle in the other alternate segment ($\angle ADB$). This covers all cases of the theorem.

---

Combining all cases, we have proved that the angle between the tangent drawn from an external point to a circle and the chord passing through the point of contact is equal to the angle in the alternate segment.

Hence Proved.

Given:

A triangle ABC with side lengths AB = $15 \text{ cm}$, BC = $14 \text{ cm}$, and AC = $13 \text{ cm}$.

A circle is inscribed in $\triangle ABC$, touching sides AB, BC, and CA at points P, Q, and R respectively.

---

To Find:

The lengths of the segments into which the sides are divided at the points of contact (AP, PB, BQ, QC, CR, AR).

---

Solution:

We know that the lengths of tangents drawn from an external point to a circle are equal.

Let the points of contact on the sides AB, BC, and CA be P, Q, and R respectively.

From vertex A, tangents AP and AR are drawn to the circle.

$AP = AR$

From vertex B, tangents BP and BQ are drawn to the circle.

$BP = BQ$

From vertex C, tangents CQ and CR are drawn to the circle.

$CQ = CR$

Let the lengths of these tangent segments be:

$AP = AR = x$

$BP = BQ = y$

$CQ = CR = z$

The lengths of the sides of the triangle are the sums of these segments:

$AB = AP + PB = x + y$

$BC = BQ + QC = y + z$

$AC = AR + RC = x + z$

Substitute the given side lengths:

We have a system of three linear equations. We can solve this system to find the values of x, y, and z.

From equation (1), we can express y in terms of x:

$y = 15 - x$

From equation (3), we can express z in terms of x:

$z = 13 - x$

Substitute these expressions for y and z into equation (2):

$(15 - x) + (13 - x) = 14$

$15 - x + 13 - x = 14$

Combine the constant terms and the x terms:

$28 - 2x = 14$

Subtract 14 from both sides and add 2x to both sides:

$28 - 14 = 2x$

$14 = 2x$

Divide by 2 to find x:

$x = \frac{14}{2}$

$x = 7$

Now substitute the value of x back into the expressions for y and z:

$y = 15 - 7 = 8$

$z = 13 - 7 = 6$

So, the lengths of the segments are:

$AP = AR = 7 \text{ cm}$

$BP = BQ = 8 \text{ cm}$

$CQ = CR = 6 \text{ cm}$

---

Answer:

The lengths of the segments into which the sides are divided at the points of contact are:

On side AB: AP = $7 \text{ cm}$ and PB = $8 \text{ cm}$.

On side BC: BQ = $8 \text{ cm}$ and QC = $6 \text{ cm}$.

On side CA: CR = $6 \text{ cm}$ and AR = $7 \text{ cm}$.

Given:

TP and TQ are tangents from an external point T to a circle with center O.

$\angle PTQ = 60^\circ$.

Radius of the circle, $OP = OQ = 8\sqrt{3} \text{ cm}$.

---

To Show:

1. OT is the angle bisector of $\angle PTQ$.

2. OT bisects PQ perpendicularly.

---

To Find:

The length of OT.

---

Proof (Part 1: OT bisects $\angle PTQ$):

Consider $\triangle OPT$ and $\triangle OQT$.

We have:

1. $OP = OQ$

2. PT = QT

3. OT = OT

Therefore, $\triangle OPT \cong \triangle OQT$ by the SSS (Side - Side - Side) congruence criterion.

By CPCTC (Corresponding Parts of Congruent Triangles), the corresponding angles are equal.

$\angle PTO = \angle QTO$

This means that OT bisects the angle $\angle PTQ$.

---

Proof (Part 2: OT bisects PQ perpendicularly):

Let the line segment OT intersect the line segment PQ at point R.

Consider $\triangle PRT$ and $\triangle QRT$.

We have:

1. PT = QT

2. $\angle PTR = \angle QTR$

3. RT = RT

Therefore, $\triangle PRT \cong \triangle QRT$ by the SAS (Side - Angle - Side) congruence criterion.

By CPCTC (Corresponding Parts of Congruent Triangles), the corresponding parts are equal.

$PR = QR$

This means that R is the midpoint of PQ. Since R lies on OT, the line segment OT bisects the line segment PQ.

Also, from CPCTC, we have:

$\angle PRT = \angle QRT$

Angles $\angle PRT$ and $\angle QRT$ form a linear pair on the straight line segment PQ.

The sum of angles in a linear pair is $180^\circ$.

So, $\angle PRT + \angle QRT = 180^\circ$.

Since $\angle PRT = \angle QRT$, we can write:

$2 \angle PRT = 180^\circ$

$\angle PRT = \frac{180^\circ}{2}$

$\angle PRT = 90^\circ$

Since $\angle PRT = 90^\circ$, the line segment RT is perpendicular to the line segment PQ. As R lies on OT, this means the line segment OT is perpendicular to the line segment PQ.

Thus, OT bisects PQ perpendicularly.

Proofs Completed.

---

Finding the length of OT:

We know that the radius through the point of contact is perpendicular to the tangent at that point.

So, the radius OP is perpendicular to the tangent PT at P.

$\angle OPT = 90^\circ$.

Consider the right-angled triangle $\triangle OPT$. It is right-angled at P.

We have the radius $OP = 8\sqrt{3} \text{ cm}$ and the angle $\angle PTO = \angle QTO = 30^\circ$ (since OT bisects $\angle PTQ = 60^\circ$).

In $\triangle OPT$, OP is the side opposite to $\angle PTO$, and OT is the hypotenuse. We can use the sine ratio:

$\sin(\angle PTO) = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{OP}{OT}$

Substitute the values:

$\sin(30^\circ) = \frac{8\sqrt{3}}{OT}$

We know that $\sin(30^\circ) = \frac{1}{2}$.

$\frac{1}{2} = \frac{8\sqrt{3}}{OT}$

Cross-multiply to find OT:

$OT \times 1 = 2 \times 8\sqrt{3}$

$OT = 16\sqrt{3}$ cm.

---

Answer:

OT is the angle bisector of $\angle PTQ$ and also bisects PQ perpendicularly.

The length of OT is $16\sqrt{3} \text{ cm}$.