Classwise Concept with Examples
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Class 10th Chapters
1. Real Numbers	2. Polynomials	3. Pair of Linear Equations in Two Variables
4. Quadratic Equations	5. Arithmetic Progressions	6. Triangles
7. Coordinate Geometry	8. Introduction to Trigonometry	9. Some Applications of Trigonometry
10. Circles	11. Constructions	12. Areas Related to Circles
13. Surface Areas And Volumes	14. Statistics	15. Probability

Content On This Page
Tangent and Secant	Properties Related to the Tangent of a Circle

Chapter 10 Circles (Concepts)

Continuing our exploration of the geometry of Circles, this chapter builds upon the foundational concepts from Class 9 and focuses specifically on the properties and behavior of lines that interact with a circle in a very particular way: the tangents. Understanding tangents is crucial not only in pure geometry but also in applications ranging from optics (how light reflects or refracts) to engineering design (like gears and belts). We will define tangents precisely and explore two fundamental theorems that govern their relationship with the circle's center, radius, and external points.

First, let's clearly define the key terms. Unlike a secant, which is a line that intersects a circle at two distinct points, a tangent to a circle is a line that touches the circle at exactly one point. This unique point where the tangent line touches the circle is called the point of contact. Imagine a wheel rolling on a straight path; the path acts as a tangent to the wheel at the point where it touches the ground.

This chapter revolves around two cornerstone theorems concerning tangents, which we will prove and apply extensively:

Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
This theorem establishes a fundamental geometric relationship: if you draw a radius to the point where a tangent touches the circle, the angle formed between the radius and the tangent line is always exactly $90^\circ$. This perpendicularity is a defining characteristic and is used constantly in solving problems.
Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.
Imagine a point $P$ located outside a circle. If we draw two tangent lines from $P$ that touch the circle at points $A$ and $B$ respectively, then the lengths of the segments $PA$ and $PB$ must be equal ($PA = PB$). The proof of this theorem typically involves connecting the center of the circle ($O$) to the external point ($P$) and to the points of contact ($A$ and $B$), forming two right-angled triangles ($\triangle OAP$ and $\triangle OBP$). These triangles are then proven congruent, usually using the RHS (Right angle-Hypotenuse-Side) congruence criterion, from which $PA=PB$ follows by CPCT.

These two theorems have several important consequences and applications that we will explore:

When two tangents are drawn from an external point $P$ to a circle with center $O$, the line segment $OP$ joining the center to the external point bisects the angle formed between the two tangents ($\angle APB$).
Furthermore, the same line segment $OP$ also bisects the angle formed at the center by the two radii to the points of contact ($\angle AOB$).

These properties arise directly from the congruent triangles established in the proof of Theorem 10.2.

Problem-solving in this chapter heavily relies on applying these theorems and their consequences. Typical problems include:

Finding unknown angles in diagrams involving tangents, radii, and chords, utilizing Theorem 10.1 (the $90^\circ$ angle) along with angle sum properties of triangles and quadrilaterals.
Calculating lengths related to tangents, often using Theorem 10.2 (equality of tangent segments) in conjunction with the Pythagorean theorem in the right-angled triangles formed.
Proving geometric properties related to quadrilaterals that circumscribe a circle (meaning all four sides are tangent to the circle). For example, proving that the opposite sides of such a quadrilateral subtend supplementary angles at the center of the circle, or demonstrating that if a parallelogram circumscribes a circle, it must necessarily be a rhombus.
Analyzing scenarios involving tangents drawn to concentric circles (circles with the same center but different radii).

This chapter, therefore, significantly deepens our understanding of circle geometry by providing a focused analysis of the unique and important properties of tangents.

Tangent and Secant of a Circle

In Class 9, you were introduced to the basic definition of a circle and various related terms such as radius, diameter, chord, arc, segment, and sector. You also learned about properties of chords and angles subtended by arcs.

In Class 10, we will further explore the geometry of circles, particularly focusing on the relationship between a circle and a straight line in the same plane. When we consider a straight line and a circle in the same plane, there are exactly three possible ways they can be positioned relative to each other in terms of intersection points:

The line does not intersect the circle at all. There are no common points between the line and the circle.
The line intersects the circle at two distinct points. This line is given a special name: a secant.
The line intersects the circle at exactly one point. This line is also given a special name: a tangent.

Secant of a Circle

A secant to a circle is a straight line that passes through the circle and intersects it at two distinct points. A chord is a line segment whose endpoints lie on the circle; it is essentially the portion of a secant that lies inside the circle and connects the two intersection points.

In the figure, line $l$ is a secant intersecting the circle at points A and B. The segment AB is a chord of the circle.

Tangent to a Circle

A tangent to a circle is a straight line in the plane of the circle that touches the circle at precisely one point. This unique point of intersection is called the point of contact or the point of tangency.

In the figure, line $m$ is a tangent to the circle, and P is its point of contact with the circle.

Tangent as a Limiting Case of a Secant:

A tangent can be understood as a limiting case of a secant. Imagine a secant line intersecting a circle at two points, P and Q. If we keep point P fixed and move point Q along the circle closer and closer to P, the secant line PQ will rotate about P. As Q gets infinitely close to P and eventually coincides with P, the secant line approaches the position of the tangent line at P.

This concept highlights that a tangent is, in a way, a secant where the two intersection points have merged into a single point of contact.

Number of Tangents from a Point to a Circle:

The number of tangents that can be drawn to a circle from a given point depends on the location of the point relative to the circle:

From a point inside the circle: No tangent can be drawn to a circle from a point that lies inside the circle. Any line passing through an interior point must intersect the circle at two distinct points.

From a point on the circle: There is exactly one tangent that can be drawn to a circle at a point that lies on the circle. The point itself serves as the point of contact for this unique tangent.

From a point outside the circle: From a point that lies outside the circle, exactly two tangents can be drawn to the circle. These two tangents will touch the circle at two distinct points of contact.

The lengths of the segments of these two tangents from the external point to their respective points of contact are equal. This is a significant property that we will discuss and prove in the next section.

Properties Related to the Tangent of a Circle

Tangents are fundamental to the geometry of circles. They have several key properties that we will explore through two important theorems. These theorems describe the relationship between a tangent and a radius, and the lengths of tangents drawn from an external point.

Theorem 10.1: Tangent-Radius Perpendicularity

Theorem: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Given: A circle with centre O and a line XY which is a tangent to the circle at a point P.

Diagram for Theorem 10.1: Circle with center O, tangent XY at point P, radius OP.

To Prove: The radius OP is perpendicular to the tangent line XY (i.e., $OP \perp XY$).

Construction: Take any point Q on the line XY, other than P, and join OQ.

Proof:

By definition, a tangent touches the circle at exactly one point. Since XY is a tangent at P, any other point on the line XY, like Q, must lie outside the circle.
If Q is outside the circle, its distance from the centre (OQ) must be greater than the radius (OP).

$OQ > OP$

(Since Q is outside the circle)

This condition holds true for every point Q on the line XY except for the point P itself. This means that OP is the shortest of all the line segments that can be drawn from the centre O to any point on the line XY.
A fundamental theorem in geometry states that the shortest distance from a point to a line is the perpendicular distance.
Since OP is the shortest distance from point O to the line XY, OP must be perpendicular to XY.

$OP \perp XY$

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

(Hence Proved)

Example 1. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Find the length of PQ.

Answer:

Solution

We are given a circle with centre O, radius $OP = 5$ cm, and the distance from the centre to the external point Q is $OQ = 12$ cm.

Diagram for example 1: Circle with center O, radius OP=5cm, tangent PQ at P, OQ=12cm. Triangle OPQ is right-angled at P.

By Theorem 10.1, the radius is perpendicular to the tangent at the point of contact. Therefore, $OP \perp PQ$, which means $\triangle OPQ$ is a right-angled triangle with the right angle at P.

The side opposite the right angle, OQ, is the hypotenuse.

Using the Pythagoras Theorem ($a^2 + b^2 = c^2$):

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

Substitute the known values:

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

$25 + PQ^2 = 144$

$PQ^2 = 144 - 25 = 119$

$PQ = \sqrt{119}$ cm

Answer: The length of PQ is $\sqrt{119}$ cm.

Theorem 10.2: Lengths of Tangents from an External Point

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

Given: A circle with centre O and a point P lying outside the circle. PA and PB are two tangents from P to the circle, touching at points A and B.

Diagram for Theorem 10.2: Circle with center O, external point P, tangents PA and PB to points A and B.

To Prove: The lengths of the tangent segments are equal, i.e., $PA = PB$.

Construction: Join OA, OB, and OP.

Proof:

We can form two triangles, $\triangle OAP$ and $\triangle OBP$. We will prove that these two triangles are congruent.

$OA = OB$ (Both are radii of the same circle).
$OP = OP$ (This side is common to both triangles).
$\angle OAP = \angle OBP = 90^\circ$ (By Theorem 10.1, the radius is perpendicular to the tangent at the point of contact).

Both triangles are right-angled triangles. We have shown that their hypotenuses (OP) are equal and one pair of corresponding sides (OA and OB) are equal. Therefore, by the RHS (Right angle-Hypotenuse-Side) congruence rule:

$\triangle OAP \cong \triangle OBP$

Since the triangles are congruent, their corresponding parts must be equal (CPCTC). Therefore:

$PA = PB$

(Hence Proved)

Important Corollaries from this proof:

Since the triangles are congruent, $\angle APO = \angle BPO$. This means the line joining the external point to the centre (OP) bisects the angle between the tangents ($\angle APB$).
Similarly, $\angle AOP = \angle BOP$. The line OP also bisects the angle subtended by the tangents at the centre ($\angle AOB$).

Example 2. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $\angle PTQ = 2 \angle OPQ$.

Answer:

Proof

Let's assume $\angle PTQ = \theta$.

Diagram for example 2: Circle with center O, external point T, tangents TP and TQ. Lines OP, OQ, OT, PQ are drawn.

In $\triangle TPQ$, we know from Theorem 10.2 that the lengths of tangents from an external point are equal.

$TP = TQ$.

Therefore, $\triangle TPQ$ is an isosceles triangle. The angles opposite the equal sides must be equal:

$\angle TPQ = \angle TQP$.

By the angle sum property of a triangle, $\angle TPQ + \angle TQP + \angle PTQ = 180^\circ$.

$2\angle TPQ + \theta = 180^\circ$

$2\angle TPQ = 180^\circ - \theta \implies \angle TPQ = 90^\circ - \frac{\theta}{2}$ ... (1)

Now, by Theorem 10.1, the radius OP is perpendicular to the tangent TP.

$\angle OPT = 90^\circ$.

From the figure, we can see that $\angle OPT = \angle OPQ + \angle TPQ$.

$\angle OPQ + \angle TPQ = 90^\circ$ ... (2)

Now substitute the expression for $\angle TPQ$ from (1) into (2):

$\angle OPQ + \left(90^\circ - \frac{\theta}{2}\right) = 90^\circ$

Subtract $90^\circ$ from both sides:

$\angle OPQ - \frac{\theta}{2} = 0 \implies \angle OPQ = \frac{\theta}{2}$

Substitute back $\theta = \angle PTQ$:

$\angle OPQ = \frac{\angle PTQ}{2}$

Multiplying by 2 gives the desired result:

$\angle PTQ = 2 \angle OPQ$

(Hence Proved)

Theorem 3: The Alternate Segment Theorem

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

Given: A circle with centre O. A tangent XY touches the circle at point P. PQ is a chord from the point of contact P. R is a point on the major arc (the alternate segment to $\angle YPQ$).

Diagram for Alternate Segment Theorem: Circle with center O, tangent XY at P, chord PQ. Point R is in the major arc.

To Prove: $\angle YPQ = \angle PRQ$

Construction: Draw the diameter PA passing through the centre O. Join A and Q.

Proof:

Since PA is a diameter and XY is the tangent at P, by Theorem 10.1, the radius OP (part of the diameter) is perpendicular to the tangent XY.

$\angle APY = 90^\circ$

From the diagram, we can write $\angle APY$ as the sum of two angles:

$\angle APQ + \angle YPQ = 90^\circ$

Rearranging this equation, we get:

$\angle YPQ = 90^\circ - \angle APQ$

... (i)

Now consider the triangle $\triangle APQ$. Since PA is a diameter, the angle subtended by the diameter at any point on the circumference is a right angle.

$\angle AQP = 90^\circ$

(Angle in a semicircle)

By the angle sum property in $\triangle APQ$:

$\angle PAQ + \angle APQ + \angle AQP = 180^\circ$

$\angle PAQ + \angle APQ + 90^\circ = 180^\circ$

Rearranging this equation, we get:

$\angle PAQ = 90^\circ - \angle APQ$

... (ii)

Comparing equations (i) and (ii), we can see that:

$\angle YPQ = \angle PAQ$

We also know that angles subtended by the same chord (PQ) in the same segment of the circle are equal. Points A and R are both on the major arc.

$\angle PAQ = \angle PRQ$

(Angles in the same segment)

Therefore, from the last two steps, we can conclude:

$\angle YPQ = \angle PRQ$

(Hence Proved)

Example 3. In the figure, PQ is the tangent to a circle at point A. If $\angle PAB = 72^\circ$ and $\angle BAC = 40^\circ$, find $\angle ABC$.

Answer:

Solution

Given: PQ is a tangent to the circle at point A. Chord AB makes an angle of $72^\circ$ with the tangent PQ (i.e., $\angle PAB = 72^\circ$). We are also given that $\angle BAC = 40^\circ$.

Diagram for example 3: A circle with tangent PQ at point A. A, B, C are points on the circle forming a triangle. Angle PAB is 72 degrees, angle BAC is 40 degrees.

To Find: The measure of $\angle ABC$.

According to the Alternate Segment Theorem, the angle between a tangent (PA) and a chord (AB) through the point of contact is equal to the angle subtended by the chord in the alternate segment.

In this case, the angle in the alternate segment to $\angle PAB$ is $\angle ACB$.

$\angle ACB = \angle PAB$

Since we are given $\angle PAB = 72^\circ$,

$\angle ACB = 72^\circ$

Now, consider the triangle $\triangle ABC$. We know two of its angles: $\angle BAC = 40^\circ$ and $\angle ACB = 72^\circ$. We can find the third angle, $\angle ABC$, using the angle sum property of a triangle.

$\angle ABC + \angle BAC + \angle ACB = 180^\circ$

Substitute the known values:

$\angle ABC + 40^\circ + 72^\circ = 180^\circ$

$\angle ABC + 112^\circ = 180^\circ$

$\angle ABC = 180^\circ - 112^\circ$

$\angle ABC = 68^\circ$

Answer: The measure of $\angle ABC$ is $68^\circ$.