---
title: "NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2"
url: https://www.swavid.com/maths/class/10/chapter/circles/ncert-solutions/exercise-10-2
dateModified: 2026-10-07T15:54:33+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2

Chapter 10: Circles. Every question from Exercise 10.2, with full working and the final answer.

Free PDF (9 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-10/swavid-ncert-solutions-class-10-maths-chapter-10-circles-1a2f8617f9.pdf

## EXERCISE 10.2

### Question 1

*1 mark · MCQ*

From a point $Q$, the length of the tangent to a circle is $24\text{ cm}$ and the distance of $Q$ from the centre is $25\text{ cm}$. The radius of the circle is

- $7\text{ cm}$
- $12\text{ cm}$
- $15\text{ cm}$
- $24.5\text{ cm}$

**Solution**

1. Let the centre of the circle be $O$ and $P$ be the point of contact, so $OP \perp OQ$.
2. In right-angled triangle $OPQ$, by Pythagoras theorem, $OQ^2 = OP^2 + PQ^2$.
3. $25^2 = OP^2 + 24^2 \implies 625 = OP^2 + 576 \implies OP^2 = 49 \implies OP = 7\text{ cm}$.
4. finalAnswer = "(A) $7\text{ cm}$"

**Answer:** (A) $7\text{ cm}$

> Common mistake: Adding instead of subtracting the squares of the given lengths.

### Question 2

*1 mark · MCQ*

In Fig. 10.11, if $TP$ and $TQ$ are the two tangents to a circle with centre $O$ so that $\angle POQ = 110^\circ$, then $\angle PTQ$ is equal to

- $60^\circ$
- $70^\circ$
- $80^\circ$
- $90^\circ$

**Solution**

1. The radius is perpendicular to the tangent at the point of contact, so $\angle OPT = 90^\circ$ and $\angle OQT = 90^\circ$.
2. In quadrilateral $POQT$, the sum of all interior angles is $360^\circ$, so $\angle PTQ + \angle POQ + \angle OPT + \angle OQT = 360^\circ$.
3. $\angle PTQ + 110^\circ + 90^\circ + 90^\circ = 360^\circ \implies \angle PTQ + 290^\circ = 360^\circ \implies \angle PTQ = 70^\circ$.
4. finalAnswer = "(B) $70^\circ$"

**Answer:** (B) $70^\circ$

> Common mistake: Assuming $\angle PTQ$ is equal to $\angle POQ$ or half of it incorrectly.

### Question 3

*1 mark · MCQ*

If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at angle of $80^\circ$, then $\angle POA$ is equal to

- $50^\circ$
- $60^\circ$
- $70^\circ$
- $80^\circ$

**Solution**

1. The line joining the external point to the centre bisects the angle between the tangents, so $\angle APO = \frac{1}{2} \times 80^\circ = 40^\circ$.
2. The radius is perpendicular to the tangent, so $\angle PAO = 90^\circ$.
3. In right-angled triangle $PAO$, $\angle POA = 180^\circ - 90^\circ - 40^\circ = 50^\circ$.
4. finalAnswer = "(A) $50^\circ$"

**Answer:** (A) $50^\circ$

> Common mistake: Subtracting the full angle of $80^\circ$ directly from $90^\circ$.

### Question 4

*3 marks · Proof*

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

**Solution**

1. Given: A circle with centre $O$ and a diameter $AB$ with tangents $PQ$ drawn at point $A$ and $RS$ drawn at point $B$.
2. To prove: $PQ \parallel RS$.
3. Since $AB$ is a diameter, $PQ$ is a tangent at $A$, so $OA \perp PQ \implies \angle OAP = 90^\circ$.
4. Similarly, $RS$ is a tangent at $B$, so $OB \perp RS \implies \angle OBS = 90^\circ$ (or $\angle OBR = 90^\circ$).
5. Thus, $\angle OAP = \angle OBS = 90^\circ$, which are alternate interior angles formed by line $AB$ intersecting $PQ$ and $RS$.
6. Therefore, $PQ \parallel RS$.
7. Hence proved.

**Answer:** The tangents drawn at the ends of a diameter of a circle are parallel.

> Common mistake: Failing to state that the alternate interior angles are equal to establish parallelism.

### Question 5

*3 marks · Proof*

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

**Solution**

1. Given: A circle with centre $O$, a tangent $XY$ at point $P$, and a perpendicular line $AB$ to $XY$ at $P$.
2. To prove: The perpendicular at $P$ passes through the centre $O$.
3. Let us assume that the perpendicular does not pass through $O$, but passes through another point $O'$ outside $O$.
4. Then, $O'P \perp XY \implies \angle O'PX = 90^\circ$.
5. Also, by Theorem 10.1, the radius through the point of contact is perpendicular to the tangent, so $OP \perp XY \implies \angle OPX = 90^\circ$.
6. From the above, $\angle O'PX = \angle OPX$, which is possible only when $O'$ coincides with $O$.
7. Therefore, the perpendicular at the point of contact to the tangent passes through the centre.
8. Hence proved.

**Answer:** The perpendicular at the point of contact to the tangent passes through the centre.

> Common mistake: Not using proof by contradiction or method of assumption.

### Question 6

*3 marks · Short answer*

The length of a tangent from a point $A$ at distance $5\text{ cm}$ from the centre of the circle is $4\text{ cm}$. Find the radius of the circle.

**Solution**

1. Given: Distance of point $A$ from the centre $O$ is $OA = 5\text{ cm}$, and the length of the tangent $AB$ is $4\text{ cm}$.
2. Formula: In right-angled triangle $OBA$, $OA^2 = OB^2 + AB^2$, where $OB$ is the radius.
3. Substitution: $5^2 = OB^2 + 4^2 \implies 25 = OB^2 + 16 \implies OB^2 = 25 - 16 = 9$.
4. Result: $OB = 3\text{ cm}$.

**Answer:** $3\text{ cm}$

> Common mistake: Treating the distance from the centre as the tangent length or vice versa.

### Question 7

*3 marks · Short answer*

Two concentric circles are of radii $5\text{ cm}$ and $3\text{ cm}$. Find the length of the chord of the larger circle which touches the smaller circle.

**Solution**

1. Let the two concentric circles have centre $O$. Let $AB$ be a chord of the larger circle of radius $5\text{ cm}$ which touches the smaller circle of radius $3\text{ cm}$ at point $P$.
2. Since $AB$ is a tangent to the smaller circle at $P$, $OP$ is perpendicular to $AB$ by Theorem 10.1, so $\angle OPA = 90^\circ$.
3. In right-angled triangle $OPA$, by Pythagoras theorem, $AP = \sqrt{OA^2 - OP^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\text{ cm}$.
4. Since the perpendicular from the centre of a circle to a chord bisects the chord, $AB = 2 \times AP = 2 \times 4\text{ cm} = 8\text{ cm}$.
5. The length of the chord is $8\text{ cm}$.

**Answer:** $8\text{ cm}$

> Common mistake: Confusing the radius of the larger circle with the distance from the centre to the chord.

### Question 8

*4 marks · Proof*

A quadrilateral $ABCD$ is drawn to circumscribe a circle (see Fig. 10.12). Prove that $AB + CD = AD + BC$

**Solution**

1. We are given a quadrilateral $ABCD$ circumscribing a circle with centre $O$, touching at points $P$, $Q$, $R$, and $S$ on sides $AB$, $BC$, $CD$, and $DA$ respectively.
2. We know that the lengths of tangents drawn from an external point to a circle are equal (Theorem 10.2).
3. Therefore, $AP = AS$, $BP = BQ$, $CR = CQ$, and $DR = DS$.
4. Adding all these equations: $(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$.
5. Simplifying the sums of segments, we get $AB + CD = AD + BC$.
6. Hence proved.

**Answer:** $AB + CD = AD + BC$

> Common mistake: Writing the vertices in wrong order while adding tangent segments.

### Question 9

*4 marks · Proof*

In Fig. 10.13, $XY$ and $X'Y'$ are two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X'Y'$ at $B$. Prove that $\angle AOB = 90^\circ$.

**Solution**

1. Let the circle have centre $O$, with parallel tangents $XY$ and $X'Y'$ at points $P$ and $Q$. Another tangent $AB$ touches the circle at $C$ and intersects $XY$ at $A$ and $X'Y'$ at $B$. Join $OC$.
2. In triangles $\triangle OPA$ and $\triangle OCA$, $OP = OC$ (radii of the same circle), $PA = CA$ (tangents from external point $A$), and $OA = OA$ (common).
3. Therefore, $\triangle OPA \cong \triangle OCA$ by SSS congruence, which gives $\angle POA = \angle COA$, so $\angle POC = 2\angle COA$.
4. Similarly, $\triangle OQB \cong \triangle OCB$, which gives $\angle QOC = 2\angle COB$.
5. Since $POQ$ is a diameter of the circle, it is a straight line, so $\angle POC + \angle QOC = 180^\circ$.
6. Thus, $2(\angle COA + \angle COB) = 180^\circ$, which means $\angle AOB = 90^\circ$.
7. Hence proved.

**Answer:** $\angle AOB = 90^\circ$

> Common mistake: Failing to establish that $POQ$ is a straight line passing through the centre.

### Question 10

*4 marks · Proof*

Prove 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 centre.

**Solution**

1. Given: A circle with centre $O$, an external point $P$, and two tangents $PA$ and $PB$ touching the circle at $A$ and $B$.
2. To prove: $\angle APB + \angle AOB = 180^\circ$.
3. Proof: By Theorem 10.1, the radius is perpendicular to the tangent at the point of contact. Therefore, $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$.
4. Consider the quadrilateral $OAPB$. The sum of all interior angles of a quadrilateral is $360^\circ$.
5. So, $\angle OAP + \angle AOB + \angle OBP + \angle APB = 360^\circ$.
6. Substituting the values, $90^\circ + \angle AOB + 90^\circ + \angle APB = 360^\circ$.
7. This gives $\angle AOB + \angle APB + 180^\circ = 360^\circ$, or $\angle APB + \angle AOB = 180^\circ$.
8. Hence proved.

**Answer:** Hence proved that the angle between the two tangents is supplementary to the angle subtended at the centre.

> Common mistake: Confusing supplementary angles with complementary angles.

### Question 11

*4 marks · Proof*

Prove that the parallelogram circumscribing a circle is a rhombus.

**Solution**

1. Given: A parallelogram $ABCD$ circumscribing a circle.
2. To prove: $ABCD$ is a rhombus.
3. Proof: Since $ABCD$ is a parallelogram, its opposite sides are equal, so $AB = CD$ and $AD = BC$.
4. From the property of tangents drawn from an external point, we know that $AB + CD = AD + BC$ (proved similarly to Question 8).
5. Since $AB = CD$ and $AD = BC$, we can write $AB + AB = AD + AD$, which gives $2AB = 2AD$, or $AB = AD$.
6. A parallelogram with adjacent sides equal is a rhombus. Therefore, $ABCD$ is a rhombus.
7. Hence proved.

**Answer:** Hence proved that the parallelogram circumscribing a circle is a rhombus.

> Common mistake: Forgetting to prove that adjacent sides are equal after proving $AB + CD = AD + BC$.

### Question 12

*3 marks · Short answer*

A triangle $ABC$ is drawn to circumscribe a circle of radius $4\text{ cm}$ such that the segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of lengths $8\text{ cm}$ and $6\text{ cm}$ respectively (see Fig. 10.14). Find the sides $AB$ and $AC$.

**Solution**

1. Let the circle touch the sides $AB$, $BC$, and $CA$ of triangle $ABC$ at points $F$, $D$, and $E$ respectively, with centre $O$ and radius $OD = 4\text{ cm}$.
2. We are given $BD = 8\text{ cm}$ and $CD = 6\text{ cm}$. Since lengths of tangents from an external point are equal, $BF = BD = 8\text{ cm}$ and $CE = CD = 6\text{ cm}$.
3. Let $AF = AE = x\text{ cm}$. Then the sides are $AB = x + 8$, $BC = 14$, and $AC = x + 6$, and the semi-perimeter $s = \frac{(x+8) + 14 + (x+6)}{2} = x + 14$.
4. The area of triangle $ABC$ can be written in two ways: $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(x+14)(6)(8)(x)} = \sqrt{48x(x+14)}$.
5. Also, $\text{Area} = \text{Area}(\triangle OAB) + \text{Area}(\triangle OBC) + \text{Area}(\triangle OCA) = \frac{1}{2}AB \times 4 + \frac{1}{2}BC \times 4 + \frac{1}{2}AC \times 4 = 2(AB + BC + AC) = 4(x + 14)$.
6. Squaring both sides of $48x(x+14) = 16(x+14)^2$ and solving for $x$, we get $3x = x + 14$, which gives $2x = 14$, so $x = 7\text{ cm}$.
7. Therefore, $AB = 7 + 8 = 15\text{ cm}$ and $AC = 7 + 6 = 13\text{ cm}$.

**Answer:** $AB = 15\text{ cm}$, $AC = 13\text{ cm}$

> Common mistake: Forgetting to equate the two expressions for the area of the triangle.

### Question 13

*5 marks · Proof*

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

**Solution**

1. Let ABCD be a quadrilateral circumscribing a circle with centre O, touching the sides AB, BC, CD and DA at points P, Q, R and S respectively.
2. Join the vertices A, B, C and D to the centre O, and also join OP, OQ, OR and OS.
3. We know that the lengths of tangents drawn from an external point to a circle are equal, and the angle between the tangents is bisected by the line joining the point to the centre.
4. Consider $\angle 1 = \angle 2$ (at vertex A), $\angle 3 = \angle 4$ (at vertex B), $\angle 5 = \angle 6$ (at vertex C), and $\angle 7 = \angle 8$ (at vertex D).
5. The sum of all angles around the centre O is $360^\circ$, so $(\angle 1 + \angle 2) + (\angle 3 + \angle 4) + (\angle 5 + \angle 6) + (\angle 7 + \angle 8) = 360^\circ$.
6. This can be written as $2(\angle 1 + \angle 8 + \angle 4 + \angle 5) = 360^\circ$ or $2(\angle AOB + \angle COD) = 360^\circ$.
7. Therefore, $\angle AOB + \angle COD = 180^\circ$, showing that opposite sides AB and CD subtend supplementary angles at the centre.
8. Similarly, it can be proved that $\angle BOC + \angle DOA = 180^\circ$. Hence proved.

**Answer:** Hence proved that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

> Common mistake: Forgetting to state that the angle between the two tangents is bisected by the line joining the external point to the centre.

## Related pages

- [All Chapter 10 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/circles/ncert-solutions)
- [Exercise 10.1](https://www.swavid.com/maths/class/10/chapter/circles/ncert-solutions/exercise-10-1)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student