Chapter 10 Circles

A circle is a collection of infinitely many points in a plane that are equidistant from a fixed point (the center).

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

For every point on the circumference of the circle, we can draw a unique tangent line passing through that point.

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

Therefore, a circle can have infinitely many tangents.

(i) A tangent to a circle intersects it in one point (s).

Explanation:

By definition, a tangent is a special line in the plane of a circle that touches the circle at exactly one point. If a line intersects the circle at two points, it is called a secant, and if it does not intersect at all, it is a non-intersecting line.

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(ii) A line intersecting a circle in two points is called a secant.

Explanation:

A secant is a line that passes through two distinct points on a circle. The line segment of the secant that lies inside the circle is called a chord of the circle.

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(iii) A circle can have two parallel tangents at the most.

Explanation:

A tangent to a circle is always perpendicular to the radius at the point of contact. For two tangents to be parallel, their points of contact must be the endpoints of a diameter of the circle. Since a diameter has only two endpoints, a circle can have only one pair of parallel tangents.

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(iv) The common point of a tangent to a circle and the circle is called point of contact.

Explanation:

This is the standard terminology. The specific point where the tangent line and the circle meet is defined as the point of contact. It is also sometimes referred to as the point of tangency.

Given:

A circle with center O.

Radius of the circle, $OP = 5$ cm.

PQ is a tangent to the circle at point P.

The distance from the center to point Q, $OQ = 12$ cm.

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To Find:

The length of the tangent, PQ.

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Solution and Justification:

We know a key theorem of circles which states that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Here, P is the point of contact, OP is the radius, and PQ is the tangent.

Therefore, the radius OP is perpendicular to the tangent PQ.

This means that $\triangle OPQ$ is a right-angled triangle with the right angle at P, i.e., $\angle OPQ = 90^\circ$.

In the right-angled triangle $\triangle OPQ$, the side opposite the right angle, OQ, is the hypotenuse.

By applying the Pythagorean theorem, we have:

Substitute the given values into the equation:

$(12)^2 = (5)^2 + PQ^2$

$144 = 25 + PQ^2$

To find the length of PQ, we rearrange the equation:

$PQ^2 = 144 - 25$

$PQ^2 = 119$

Taking the square root of both sides:

$PQ = \sqrt{119}$ cm.

The length of PQ is $\sqrt{119}$ cm. This matches option (D).

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Conclusion:

The correct option is (D) $\sqrt{119}$ cm.

To perform this construction, we follow these steps:

Construction Steps:

1. First, draw a circle with a center point, let's call it O.

2. Draw a line anywhere on the plane, which will be our "given line". Let's call this line L.

3. Now, to create the parallel tangent, draw a line from the center O that is perpendicular to the given line L. This perpendicular line will intersect the circle at two points on opposite sides. Let's call one of these intersection points P.

4. Draw a line through point P that is parallel to the given line L. This new line is a tangent to the circle because it touches the circle at exactly one point (P) and is perpendicular to the radius OP.

5. To create the parallel secant, choose any point Q on the radius OP (or anywhere inside the circle on the perpendicular line). Draw a line through point Q that is parallel to the given line L. This line will intersect the circle at two distinct points, making it a secant to the circle.

The resulting figure shows the original circle, the given line L, a tangent line parallel to L, and a secant line also parallel to L.

Given:

Two concentric circles, $C_1$ (larger) and $C_2$ (smaller), with a common center O.

AB is a chord of the larger circle $C_1$ that is tangent to the smaller circle $C_2$ at point P.

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To Prove:

The chord AB is bisected at the point of contact P, i.e., AP = PB.

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Construction:

Join OP.

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Proof:

Consider the smaller circle $C_2$.

AB is a tangent to the circle $C_2$ at the point P.

OP is the radius of the smaller circle $C_2$ through the point of contact P.

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

Now, consider the larger circle $C_1$.

AB is a chord of the circle $C_1$.

OP is a line segment from the center O such that $OP \perp AB$.

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

Therefore, OP bisects the chord AB.

This implies that P is the midpoint of AB.

Hence, AP = PB.

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Thus, it is proved that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

Given:

A circle with centre O.

An external point T.

TP and TQ are two tangents drawn from T to the circle at points P and Q respectively.

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To Prove:

$\angle PTQ = 2 \angle OPQ$.

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Proof:

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

In triangle TPQ, since TP = TQ, it is an isosceles triangle.

Therefore, the angles opposite to equal sides are equal.

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

So, $\angle OPT = 90^\circ$.

From the figure, we can write $\angle OPT$ as the sum of $\angle OPQ$ and $\angle TPQ$.

$\angle OPT = \angle OPQ + \angle TPQ$

Rearranging this equation to find $\angle TPQ$:

Now, consider the angle sum property of triangle TPQ:

$\angle PTQ + \angle TPQ + \angle TQP = 180^\circ$

Since $\angle TPQ = \angle TQP$, we can write:

$\angle PTQ + \angle TPQ + \angle TPQ = 180^\circ$

Substitute the expression for $\angle TPQ$ from equation (i) into equation (ii):

$\angle PTQ + 2 (90^\circ - \angle OPQ) = 180^\circ$

$\angle PTQ + 180^\circ - 2 \angle OPQ = 180^\circ$

Subtract $180^\circ$ from both sides:

$\angle PTQ - 2 \angle OPQ = 0$

$\angle PTQ = 2 \angle OPQ$

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Hence Proved.

Given:

A circle with centre O and radius OP = OQ = 5 cm.

PQ is a chord of length 8 cm.

Tangents at P and Q intersect at point T.

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To Find:

The length of the tangent TP.

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Construction:

Join OT. Let OT intersect the chord PQ at point R.

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Solution:

We know that the line joining the centre of a circle to the external point from which tangents are drawn bisects the angle between the tangents and is the perpendicular bisector of the chord joining the points of contact.

Therefore, OT is the perpendicular bisector of PQ.

So, OR $\perp$ PQ and R is the midpoint of PQ.

Now, consider the right-angled triangle $\triangle ORP$ (right-angled at R).

By the Pythagorean theorem:

$OR^2 + PR^2 = OP^2$

Substitute the known values:

$OR^2 + 16 = 25$

$OR^2 = 25 - 16 = 9$

$OR = \sqrt{9} = 3$ cm (Since length must be positive)

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

So, $\triangle OPT$ is a right-angled triangle with the right angle at P.

Let TP = $x$ and TR = $y$.

In right-angled $\triangle PRT$ (right-angled at R):

$TP^2 = PR^2 + TR^2$

In right-angled $\triangle OPT$ (right-angled at P):

$OT^2 = OP^2 + TP^2$

From the figure, $OT = OR + TR = 3 + y$.

Substitute the values:

Substitute the value of $x^2$ from equation (i) into equation (ii):

$9 + 6y + y^2 = 25 + (16 + y^2)$

$9 + 6y + y^2 = 41 + y^2$

Subtract $y^2$ from both sides:

$9 + 6y = 41$

$6y = 41 - 9$

$6y = 32$

$y = \frac{32}{6} = \frac{16}{3}$ cm.

Now, substitute the value of $y$ back into equation (i) to find $x$ (which is TP):

$x^2 = 16 + y^2$

$x^2 = 16 + \left(\frac{16}{3}\right)^2$

$x^2 = 16 + \frac{256}{9}$

$x^2 = \frac{16 \times 9 + 256}{9}$

$x^2 = \frac{144 + 256}{9}$

$x^2 = \frac{400}{9}$

Take the square root:

$x = \sqrt{\frac{400}{9}} = \frac{20}{3}$

So, the length TP = $\frac{20}{3}$ cm.

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Alternate Solution (Using Similar Triangles):

In $\triangle TRP$, $\angle TRP = 90^\circ$.

In $\triangle PRO$, $\angle PRO = 90^\circ$.

We know OP $\perp$ TP, so $\angle OPT = 90^\circ$.

$\angle OPR + \angle TPR = \angle OPT = 90^\circ$.

In right-angled $\triangle PRO$, $\angle POR + \angle OPR = 90^\circ$.

From these two equations, we have $\angle POR = \angle TPR$.

Now, consider $\triangle TRP$ and $\triangle PRO$.

Therefore, by AA similarity criterion:

$\triangle TRP \sim \triangle PRO$

The corresponding sides are proportional:

$\frac{TR}{PR} = \frac{RP}{RO} = \frac{TP}{PO}$

We know PR = 4 cm, RO = 3 cm, and PO = 5 cm (radius).

Using the ratio $\frac{RP}{RO} = \frac{TP}{PO}$:

Solving for TP:

$TP = \frac{4 \times 5}{3} = \frac{20}{3}$ cm.

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Therefore, the length of TP is $\frac{20}{3}$ cm.

Given:

Let the circle have its center at O.

Let Q be an external point and P be the point of contact on the circle.

Length of the tangent from Q, $PQ = 24$ cm.

Distance of Q from the center, $OQ = 25$ cm.

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To Find:

The radius of the circle, OP.

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Justification:

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

Therefore, $OP \perp PQ$.

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

By the Pythagorean theorem, the square of the hypotenuse (OQ) is equal to the sum of the squares of the other two sides (OP and PQ).

Substitute the given values:

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

$625 = OP^2 + 576$

Rearranging the equation to solve for the radius, OP:

$OP^2 = 625 - 576$

$OP^2 = 49$

$OP = \sqrt{49}$

$OP = 7$ cm

The radius of the circle is 7 cm.

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Conclusion:

The correct option is (A) 7 cm.

Given:

A circle with centre O.

TP and TQ are tangents to the circle from an external point T.

The angle at the center, $\angle POQ = 110^\circ$.

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To Find:

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

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Justification:

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

Therefore, the radius OP is perpendicular to the tangent TP, and the radius OQ is perpendicular to the tangent TQ.

Now, consider the quadrilateral OPTQ. The sum of the interior angles of a quadrilateral is $360^\circ$.

$\angle OPT + \angle POQ + \angle OQT + \angle PTQ = 360^\circ$

Substitute the known angle measures into the equation:

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

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

Solve for $\angle PTQ$:

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

$\angle PTQ = 70^\circ$

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Conclusion:

The correct option is (B) 70°.

Given:

A circle with centre O.

PA and PB are tangents from an external point P.

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

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To Find:

The measure of $\angle POA$.

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Justification:

Consider the triangles $\triangle OAP$ and $\triangle OBP$.

By the SSS (Side-Side-Side) congruence criterion, $\triangle OAP \cong \triangle OBP$.

By CPCT (Corresponding Parts of Congruent Triangles), the corresponding angles are equal.

This means the line OP bisects the angle $\angle APB$.

$\angle OPA = \frac{1}{2} \angle APB = \frac{1}{2} \times 80^\circ = 40^\circ$.

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

So, $\angle OAP = 90^\circ$.

Now, in the right-angled triangle $\triangle OAP$, the sum of angles is $180^\circ$.

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

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

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

$\angle POA = 180^\circ - 130^\circ = 50^\circ$.

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Conclusion:

The correct option is (A) 50°.

Given:

A circle with centre O and diameter PQ.

Two tangents, AB and CD, are drawn at the endpoints of the diameter, P and Q, respectively.

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To Prove:

The tangents are parallel, i.e., $AB \parallel CD$.

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Proof:

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

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

Similarly, since CD is the tangent at point Q, the radius OQ is perpendicular to CD.

The diameter PQ is a straight line. Let's consider PQ as a transversal intersecting the lines AB and CD.

The angles $\angle OPA$ (which is the same as $\angle QPA$) and $\angle OQD$ (which is the same as $\angle PQC$) are alternate interior angles.

Let's consider $\angle APQ$ and $\angle PQC$.

Since $\angle APQ = \angle PQC = 90^\circ$, this pair of alternate interior angles are equal.

When the alternate interior angles formed by a transversal cutting two lines are equal, the lines are parallel.

Therefore, $AB \parallel CD$.

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Hence, the tangents drawn at the ends of a diameter of a circle are parallel.

Given:

A circle with centre O.

A tangent line XY touching the circle at the point of contact P.

A line segment ZP is drawn perpendicular to XY at P.

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To Prove:

The perpendicular ZP passes through the centre O.

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Proof:

We will use the method of proof by contradiction.

Let's assume that the perpendicular ZP at the point of contact P does not pass through the centre O.

If it does not pass through O, let us join the centre O to the point of contact P. The line segment OP is the radius of the circle.

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

Therefore, $OP \perp XY$.

From our initial setup, we are given that the line ZP is perpendicular to the tangent XY at P.

From these two statements, we have:

$\angle OPX = \angle ZPX = 90^\circ$.

This is only possible if the line segments OP and ZP coincide, meaning they are the same line. For this to happen, the point O must lie on the line ZP.

This contradicts our initial assumption that the perpendicular ZP does not pass through the centre O.

Therefore, our assumption was incorrect.

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Hence, the perpendicular at the point of contact to the tangent to a circle must pass through the centre.

Given:

A circle with centre O.

A point A is at a distance of 5 cm from the centre, so $OA = 5$ cm.

The length of the tangent from point A to the circle is 4 cm. Let P be the point of contact, so $AP = 4$ cm.

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To Find:

The radius of the circle, OP.

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Solution:

The radius of a circle is perpendicular to the tangent at the point of contact.

Therefore, $OP \perp AP$, which means $\triangle OPA$ is a right-angled triangle with the right angle at P.

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

Substitute the given values:

$(5)^2 = OP^2 + (4)^2$

$25 = OP^2 + 16$

Solve for $OP^2$:

$OP^2 = 25 - 16 = 9$

Take the square root to find OP:

$OP = \sqrt{9} = 3$

The radius of the circle is 3 cm.

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Answer: The radius of the circle is 3 cm.

Given:

Two concentric circles with a common centre O.

Radius of the larger circle, $R = 5$ cm.

Radius of the smaller circle, $r = 3$ cm.

AB is a chord of the larger circle that is tangent to the smaller circle at point P.

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To Find:

The length of the chord AB.

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Solution:

Join OA and OP. Here, OA is the radius of the larger circle and OP is the radius of the smaller circle.

$OA = 5$ cm and $OP = 3$ cm.

Since AB is a tangent to the smaller circle at point P, the radius OP must be perpendicular to the tangent AB.

Therefore, $\angle OPA = 90^\circ$, and $\triangle OPA$ is a right-angled triangle.

By the Pythagorean theorem in $\triangle OPA$:

Substitute the known values:

$(5)^2 = (3)^2 + AP^2$

$25 = 9 + AP^2$

$AP^2 = 25 - 9 = 16$

$AP = \sqrt{16} = 4$ cm.

Now, consider the chord AB of the larger circle. The line segment OP is perpendicular from the center O to the chord AB. A perpendicular from the center of a circle to a chord bisects the chord.

Therefore, P is the midpoint of AB, and $AB = 2 \times AP$.

$AB = 2 \times 4$ cm = $8$ cm.

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Answer: The length of the chord of the larger circle is 8 cm.

Given:

A quadrilateral ABCD circumscribes a circle with center O. The sides AB, BC, CD, and DA touch the circle at points P, Q, R, and S respectively.

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To Prove:

$AB + CD = AD + BC$.

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Proof:

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

Applying this theorem for the external points A, B, C, and D:

Now, let's add these four equations:

$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$

From the figure, we can group the segments to form the sides of the quadrilateral:

$AP + BP = AB$

$CR + DR = CD$

$AS + DS = AD$

$BQ + CQ = BC$

Substituting these into the summed equation:

$AB + CD = AD + BC$

Hence Proved.

Given:

A circle with centre O.

$XY \parallel X'Y'$ are two tangents to the circle at points P and Q respectively.

A third tangent AB touches the circle at C and intersects XY at A and X'Y' at B.

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To Prove:

$\angle AOB = 90^\circ$.

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Construction:

Join OC.

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Proof:

Consider the external point A. AP and AC are tangents from A to the circle.

In $\triangle OPA$ and $\triangle OCA$:

By SSS congruence, $\triangle OPA \cong \triangle OCA$.

Therefore, by CPCT, $\angle POA = \angle COA$. ... (i)

Similarly, consider the external point B. BQ and BC are tangents from B.

In $\triangle OQB$ and $\triangle OCB$:

By SSS congruence, $\triangle OQB \cong \triangle OCB$.

Therefore, by CPCT, $\angle QOB = \angle COB$. ... (ii)

Since XY and X'Y' are parallel tangents, PQ is a diameter and thus a straight line.

The sum of angles on the straight line PQ at the center O is $180^\circ$.

$\angle POA + \angle COA + \angle COB + \angle QOB = 180^\circ$

Using (i) and (ii), we can substitute $\angle POA$ with $\angle COA$ and $\angle QOB$ with $\angle COB$:

$\angle COA + \angle COA + \angle COB + \angle COB = 180^\circ$

$2\angle COA + 2\angle COB = 180^\circ$

$2(\angle COA + \angle COB) = 180^\circ$

$\angle COA + \angle COB = 90^\circ$

From the figure, $\angle AOB = \angle COA + \angle COB$.

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

Hence Proved.

Given:

A circle with centre O.

An external point P from which two tangents PA and PB are drawn to the circle.

A and B are the points of contact.

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To Prove:

The angle between the tangents ($\angle APB$) and the angle subtended by the line segment joining the points of contact at the centre ($\angle AOB$) are supplementary.

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

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Proof:

The radius of a circle is perpendicular to the tangent at the point of contact.

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

Now consider the quadrilateral OAPB. The sum of the interior angles of a quadrilateral is $360^\circ$.

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

Substitute the known values:

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

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

Subtract $180^\circ$ from both sides:

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

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

Since the sum of the two angles is $180^\circ$, they are supplementary.

Hence Proved.

Given:

A parallelogram ABCD circumscribes a circle with center O.

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

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To Prove:

The parallelogram ABCD is a rhombus.

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Proof:

Since ABCD is a parallelogram, its opposite sides are equal.

For any quadrilateral circumscribing a circle, the sum of opposite sides are equal (as proved in Question 8).

Therefore, for quadrilateral ABCD:

Now, substitute the properties of the parallelogram from equations (i) and (ii) into equation (iii).

Replace CD with AB and DA with BC:

$AB + AB = BC + BC$

$2AB = 2BC$

$AB = BC$

This shows that two adjacent sides of the parallelogram are equal.

Since opposite sides are equal and we have now shown that adjacent sides are also equal, all four sides must be equal.

$AB = BC = CD = DA$.

A parallelogram with all sides equal is a rhombus.

Therefore, ABCD is a rhombus.

Hence Proved.

Given:

A triangle ABC circumscribes a circle with centre O and radius $r = 4$ cm.

The point of contact D divides the side BC into segments $BD = 8$ cm and $DC = 6$ cm.

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To Find:

The lengths of the sides AB and AC.

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Solution:

Let the circle touch the sides AB at point E and AC at point F.

We know that tangents from an external point to a circle are equal in length.

From vertex B: $BE = BD = 8$ cm.

From vertex C: $CF = CD = 6$ cm.

From vertex A: Let $AE = AF = x$ cm.

The lengths of the sides of $\triangle ABC$ are:

$a = BC = BD + DC = 8 + 6 = 14$ cm.

$b = AC = AF + FC = x + 6$ cm.

$c = AB = AE + EB = x + 8$ cm.

We will find the area of the triangle using two different methods and equate them to find $x$.

Method 1: Area using Heron's Formula

Semi-perimeter, $s = \frac{a+b+c}{2} = \frac{14 + (x+6) + (x+8)}{2} = \frac{2x+28}{2} = x+14$.

Area$(\triangle ABC) = \sqrt{s(s-a)(s-b)(s-c)}$

Area$(\triangle ABC) = \sqrt{(x+14)( (x+14)-14 )( (x+14)-(x+6) )( (x+14)-(x+8) )}$

Area$(\triangle ABC) = \sqrt{(x+14)(x)(8)(6)} = \sqrt{48x(x+14)}$

Method 2: Area as sum of smaller triangles

Area$(\triangle ABC) = \text{Area}(\triangle OBC) + \text{Area}(\triangle OCA) + \text{Area}(\triangle OAB)$

The radius is the height for each of these triangles with bases being the sides of $\triangle ABC$. Radius $r=4$ cm.

Area$(\triangle ABC) = \frac{1}{2} \times BC \times r + \frac{1}{2} \times AC \times r + \frac{1}{2} \times AB \times r$

Area$(\triangle ABC) = \frac{1}{2} \times r \times (BC + AC + AB) = r \times s$

Area$(\triangle ABC) = 4 \times (x+14)$

Equating the two areas:

$\sqrt{48x(x+14)} = 4(x+14)$

Square both sides:

$48x(x+14) = 16(x+14)^2$

Since $x+14 \neq 0$, we can divide both sides by $16(x+14)$:

$\frac{48x}{16} = x+14$

$3x = x+14$

$2x = 14$

$x = 7$ cm.

Now, we can find the lengths of the sides AB and AC.

$AB = x + 8 = 7 + 8 = 15$ cm.

$AC = x + 6 = 7 + 6 = 13$ cm.

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Answer: The lengths of the sides are AB = 15 cm and AC = 13 cm.

Given:

A quadrilateral ABCD circumscribes a circle with centre O.

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

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To Prove:

Opposite sides subtend supplementary angles at the centre, i.e.,

$\angle AOB + \angle COD = 180^\circ$ and $\angle BOC + \angle DOA = 180^\circ$.

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Construction:

Join the center O to the vertices A, B, C, D and to the points of contact P, Q, R, S.

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Proof:

Consider the triangles $\triangle OAS$ and $\triangle OAP$.

By SSS congruence, $\triangle OAS \cong \triangle OAP$.

Therefore, by CPCT, $\angle SOA = \angle POA$. Let this be $\angle 1$. So $\angle 1 = \angle SOA = \angle POA$.

Similarly, we can prove the congruence of other pairs of triangles:

$\triangle OPB \cong \triangle OQB \implies \angle POB = \angle QOB$ (Let this be $\angle 2$)

$\triangle OQC \cong \triangle ORC \implies \angle QOC = \angle ROC$ (Let this be $\angle 3$)

$\triangle ORD \cong \triangle OSD \implies \angle ROD = \angle SOD$ (Let this be $\angle 4$)

The sum of all angles around the point O is $360^\circ$.

$(\angle SOA + \angle POA) + (\angle POB + \angle QOB) + (\angle QOC + \angle ROC) + (\angle ROD + \angle SOD) = 360^\circ$

$2\angle 1 + 2\angle 2 + 2\angle 3 + 2\angle 4 = 360^\circ$

$2(\angle 1 + \angle 2 + \angle 3 + \angle 4) = 360^\circ$

$\angle 1 + \angle 2 + \angle 3 + \angle 4 = 180^\circ$

Now, let's consider the angles subtended by the opposite sides at the center.

$\angle AOB = \angle POA + \angle POB = \angle 1 + \angle 2$

$\angle COD = \angle ROC + \angle ROD = \angle 3 + \angle 4$

Summing these two angles:

$\angle AOB + \angle COD = (\angle 1 + \angle 2) + (\angle 3 + \angle 4)$

From our earlier result, we know $\angle 1 + \angle 2 + \angle 3 + \angle 4 = 180^\circ$.

Therefore, $\angle AOB + \angle COD = 180^\circ$.

Similarly,

$\angle BOC = \angle QOC + \angle QOB = \angle 3 + \angle 2$

$\angle DOA = \angle SOD + \angle SOA = \angle 4 + \angle 1$

Summing these two angles:

$\angle BOC + \angle DOA = (\angle 3 + \angle 2) + (\angle 4 + \angle 1) = \angle 1 + \angle 2 + \angle 3 + \angle 4 = 180^\circ$.

Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre.

Hence Proved.