---
title: "NCERT Solutions for Class 10 Maths Chapter 10 Circles (2026-27)"
url: https://www.swavid.com/maths/class/10/chapter/circles/ncert-solutions
dateModified: 2026-10-07T15:54:33+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 10 Circles (2026-27)

This chapter's questions cover fundamental concepts about tangents to a circle, including their existence, properties, and the lengths of tangents drawn from an external point.

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 10.1

### Question 1

*2 marks · Very short answer*

How many tangents can a circle have?

**Solution**

1. A circle is made up of infinitely many points.
2. Since there is one tangent at each point on the circle, a circle can have infinitely many tangents.

**Answer:** Infinitely many

> Common mistake: Writing finite numbers like two or four.

### Question 2

*1 mark · Fill in the blank*

Fill in the blanks :
(i) A tangent to a circle intersects it in _______ point (s).
(ii) A line intersecting a circle in two points is called a _______.
(iii) A circle can have _______ parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called _______.

**Part (i)**

1. A tangent is defined as a line that intersects the circle at only one point.

Answer (i): one

**Part (ii)**

1. A line intersecting a circle in two points is called a secant of the circle.

Answer (ii): secant

**Part (iii)**

1. A circle can have at most two parallel tangents to a given secant or parallel to each other.

Answer (iii): two

**Part (iv)**

1. The common point of a tangent to a circle and the circle is called the point of contact.

Answer (iv): point of contact

**Answer:** (i) one (ii) secant (iii) two (iv) point of contact

> Common mistake: Confusing the number of parallel tangents with the infinite number of tangents that can be drawn on a circle.

### Question 3

*1 mark · MCQ*

A tangent $PQ$ at a point $P$ of a circle of radius $5\text{ cm}$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12\text{ cm}$. Length $PQ$ is :

- $12\text{ cm}$
- $13\text{ cm}$
- $8.5\text{ cm}$
- $\sqrt{119}\text{ cm}$

**Solution**

1. Since the tangent at any point of a circle is perpendicular to the radius through the point of contact, triangle OPQ is a right-angled triangle at P.
2. Using Pythagoras theorem, $OQ^2 = OP^2 + PQ^2$, which gives $12^2 = 5^2 + PQ^2$, so $PQ = \sqrt{144 - 25} = \sqrt{119}\text{ cm}$.

**Answer:** (D) $\sqrt{119}\text{ cm}$

> Common mistake: Adding the squares instead of subtracting, leading to $13\text{ cm}$.

### Question 4

*3 marks · Short answer*

Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

**Solution**

1. Draw a given line \(l\) and a circle in a plane.
2. Draw a line \(m\) parallel to \(l\) such that it intersects the circle at only one point, making it a tangent.
3. Draw another line \(n\) parallel to \(l\) such that it intersects the circle at two points, making it a secant.

**Answer:** A circle with one tangent and one secant parallel to a given line.

> Common mistake: Drawing lines that are not parallel to the given line.

## 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.

## Frequently asked questions

### How many exercises and questions are there in NCERT Solutions for Class 10 Maths Chapter 10 Circles?

This chapter has a total of 2 exercises with 17 questions in the 2026-27 NCERT textbook. Exercise 10.1 contains 4 questions, while Exercise 10.2 contains 13 questions. You can find step-by-step solutions for all these questions in SwaVid's free PDF available on this page only.

### What are the main topics covered in Class 10 Maths Chapter 10 Circles exercises?

Exercise 10.1 covers basic terminology of circles, secants, tangents, the Pythagoras theorem applied to radius and tangent, tangent and secant parallel to a line, and tangents to a circle. Exercise 10.2 covers the angle sum property of quadrilaterals, perpendicular to tangent passes through centre, properties of tangents from an external point, radius and tangent property, radius using tangent length, tangents at the ends of a diameter, and supplementary angles subtended at the centre.

### Which is the hardest question type in this chapter and how should I approach it?

Proof-based questions in Exercise 10.2 are generally considered the most challenging by students. To approach them, you should clearly state the given information, identify the relevant theorems like the equality of tangents from an external point, and apply geometric properties logically. SwaVid's detailed solutions on this page break down these proofs into simple steps.

### How can I write answers to score full marks in board exams for Class 10 Circles?

To score full marks, always draw a neat and labelled diagram for every geometrical problem. Clearly mention the theorems you are using, such as $\text{tangent} \perp \text{radius}$, and write down every intermediate step logically. Referring to SwaVid's step-by-step solutions on this page will help you learn the correct presentation style.

### Is a free PDF available for Class 10 Maths Chapter 10 Circles solutions?

Yes, a comprehensive and free PDF containing complete solutions for all questions in this chapter is available on this page only. These solutions are prepared according to the 2026-27 NCERT textbook guidelines to help with your exam preparation.

## Related pages

- [Exercise 10.1 solutions](https://www.swavid.com/maths/class/10/chapter/circles/ncert-solutions/exercise-10-1)
- [Exercise 10.2 solutions](https://www.swavid.com/maths/class/10/chapter/circles/ncert-solutions/exercise-10-2)
- [Circles: CBSE previous year questions](https://www.swavid.com/cbse/class-10/maths/pyq/circles)
- [Class 10 Maths chapters](https://www.swavid.com/maths/class/10)

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