---
title: "Circles: CBSE Class 10 Maths previous year questions"
url: https://www.swavid.com/cbse/class-10/maths/pyq/circles
---

# Circles: CBSE Class 10 Maths previous year questions

42 questions from CBSE Class 10 board papers, newest first, each with full working.

## CBSE Class 10 Maths Standard Question Paper 2026 (Set 30/1/1) with Solutions

### Question 12

*1 mark · MCQ*

If $\text{TP}$ and $\text{TQ}$ are two tangents to a circle with centre $\text{O}$ from an external point $\text{T}$ so that $\angle \text{POQ} = 120^\circ$, then $\angle \text{PTQ}$ is equal to :

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

**Solution**

1. The sum of angles opposite to each other in quadrilateral PAOR is $180^\circ$, where $\angle OPA = \angle OQA = 90^\circ$.
2. Therefore, $\angle \text{PTQ} = 180^\circ - \angle \text{POQ} = 180^\circ - 120^\circ = 60^\circ$.

**Answer:** (a) $60^\circ$

> Common mistake: Assuming $\angle \text{PTQ}$ is equal to $\angle \text{POQ}$ or half of it without geometric proof.

### Question 13

*1 mark · MCQ*

In the given figure, $\text{PA}$ is a tangent from an external point $\text{P}$ to a circle with centre $\text{O}$. If $\angle \text{POB} = 125^\circ$, then $\angle \text{APO}$ is equal to :

- 25^\circ
- 65^\circ
- 90^\circ
- 35^\circ

**Solution**

1. Radius $OA$ is perpendicular to tangent $PA$, so $\angle OAP = 90^\circ$.
2. The angle $\angle AOB = 180^\circ - \angle POB = 180^\circ - 125^\circ = 55^\circ$.
3. In triangle $OAP$, the sum of angles is $180^\circ$, so $\angle APO = 180^\circ - 90^\circ - 55^\circ = 35^\circ$.

**Answer:** (d) 35^\circ

> Common mistake: Assuming triangle OAP is isosceles without checking side lengths.

### Question 25

*2 marks · Very short answer*

Two concentric circles are of radii $5\text{ cm}$ and $4\text{ cm}$. Find the length of the chord of the larger circle which touches the smaller circle.

**Solution**

1. Let the concentric circles be with center $O$. Let $AB$ be the chord of the larger circle of radius $5\text{ cm}$ which touches the smaller circle of radius $4\text{ cm}$ at point $P$.
2. Since $AB$ is a tangent to the smaller circle at $P$, $OP \perp AB$.
3. In right-angled triangle $\triangle OPA$, $\angle OPA = 90^\circ$, $OA = 5\text{ cm}$, and $OP = 4\text{ cm}$.
4. By Pythagoras theorem, $AP = \sqrt{OA^2 - OP^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3\text{ cm}$.
5. Since the perpendicular from the center to a chord bisects the chord, $AB = 2AP = 2 \times 3\text{ cm} = 6\text{ cm}$.

**Answer:** $6\text{ cm}$

> Common mistake: Students often forget to multiply the length of the segment $AP$ by 2 to get the total length of the chord $AB$.

### Question 29

*3 marks · Short answer*

(A) In the given figure, $\Delta \text{ABC}$ is a right triangle in which $\angle \text{B} = 90^\circ$, $\text{AB} = 4\text{ cm}$ and $\text{BC} = 3\text{ cm}$. Find the radius of the circle inscribed in the triangle $\text{ABC}$.

**Solution**

1. Given $\text{AB} = 4\text{ cm}$ and $\text{BC} = 3\text{ cm}$, find the hypotenuse $\text{AC}$ using Pythagoras theorem: $\text{AC} = \sqrt{4^2 + 3^2} = 5\text{ cm}$.
2. Let $r$ be the radius of the inscribed circle. The circle touches $\text{AB}$ and $\text{BC}$ at points making squares with the vertex $\text{B}$, so the tangents from $\text{B}$ have length $r$.
3. The lengths of tangents from vertex $\text{A}$ are $4 - r$ and from vertex $\text{C}$ are $3 - r$.
4. The sum of the lengths of tangents from $\text{C}$ and $\text{A}$ equals $\text{AC}$: $(4 - r) + (3 - r) = 5$.
5. Simplify the equation: $7 - 2r = 5$, which gives $2r = 2$, so $r = 1\text{ cm}$.
6. Thus, the radius of the inscribed circle is $1\text{ cm}$.

**Answer:** $1\text{ cm}$

> Common mistake: Confusing the radius with the side lengths of the formed square at vertex B.

### Question 29 (OR)

*3 marks · Proof*

(B) In the given figure, if a circle touches the side $\text{QR}$ of $\Delta \text{PQR}$ at $\text{S}$ and extended sides $\text{PQ}$ and $\text{PR}$ at $\text{M}$ and $\text{N}$ respectively, then prove that : $\text{PM} = \frac{1}{2}(\text{PQ} + \text{QR} + \text{PR})$

**Solution**

1. We know that the lengths of tangents drawn from an external point to a circle are equal.
2. Therefore, from point $\text{P}$, tangents are $\text{PM} = \text{PN}$.
3. From point $\text{Q}$, tangents are $\text{QM} = \text{QS}$.
4. From point $\text{R}$, tangents are $\text{RN} = \text{RS}$.
5. Consider the RHS: $\text{PQ} + \text{QR} + \text{PR} = (\text{PM} - \text{QM}) + (\text{QS} + \text{RS}) + (\text{PN} - \text{RN})$.
6. Substitute $\text{QM} = \text{QS}$ and $\text{RN} = \text{RS}$: $\text{PQ} + \text{QR} + \text{PR} = (\text{PM} - \text{QS}) + (\text{QS} + \text{RS}) + (\text{PN} - \text{RS})$.
7. Cancel out the terms to get: $\text{PQ} + \text{QR} + \text{PR} = \text{PM} + \text{PN}$.
8. Since $\text{PM} = \text{PN}$, we can write $\text{PQ} + \text{QR} + \text{PR} = 2\text{PM}$.
9. Hence, $\text{PM} = \frac{1}{2}(\text{PQ} + \text{QR} + \text{PR})$.
10. Hence proved.

**Answer:** Hence proved.

> Common mistake: Wrong substitution of extended segments in terms of tangents.

## CBSE Class 10 Maths Basic Question Paper 2025 (Set 430/1/1) with Solutions

### Question 9

*1 mark · MCQ*

In the given figure, PA is a tangent to a circle with centre O. If OP = 10 cm, then the length of AP is :

- $10\sqrt{3}$ cm
- 20 cm
- 5 cm
- $5\sqrt{3}$ cm

**Solution**

1. In right-angled triangle $OAP$, $\angle OAP = 90^{\circ}$ because the tangent at any point of a circle is perpendicular to the radius through the point of contact.
2. Using trigonometry in $\triangle OAP$, $\cos 30^{\circ} = \frac{AP}{OP}$.
3. Substitute the values: $\frac{\sqrt{3}}{2} = \frac{AP}{10}$, which gives $AP = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}~\text{cm}$.

**Answer:** (D) $5\sqrt{3}$ cm

> Common mistake: Using sine instead of cosine for the adjacent side.

### Question 23

*2 marks · Very short answer*

Two concentric circles are of radii 6 cm and 10 cm. Find the length of the chord of the larger circle which touches the smaller circle.

**Solution**

1. Let the concentric circles have center $O$. Let $AB$ be a chord of the larger circle of radius $10\text{ cm}$ which touches the smaller circle of radius $6\text{ cm}$ at point $P$.
2. The radius $OP$ is perpendicular to the chord $AB$ at the point of contact $P$.
3. In right-angled triangle $OPA$, $OA = 10\text{ cm}$ and $OP = 6\text{ cm}$.
4. By Pythagoras theorem, $AP = \sqrt{OA^2 - OP^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8\text{ cm}$.
5. The length of the chord $AB = 2 \times AP = 2 \times 8 = 16\text{ cm}$.

**Answer:** $16\text{ cm}$

> Common mistake: Forgetting to double the length of AP to get the full chord length.

### Question 29

*3 marks · Proof*

Prove that a rectangle circumscribing a circle is a square.

**Solution**

1. Let ABCD be a rectangle circumscribing a circle with centre O.
2. We know that the lengths of tangents drawn from an external point to a circle are equal.
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. This simplifies to $AB + CD = AD + BC$.
6. Since ABCD is a rectangle, $AB = CD$ and $BC = AD$. Substituting these, $2AB = 2AD$, which means $AB = AD$.
7. Thus, the adjacent sides of the rectangle are equal, making ABCD a square. Hence proved.

**Answer:** Hence proved that the rectangle is a square.

> Common mistake: Not stating the theorem about tangents from an external point clearly.

## CBSE Class 10 Maths Standard Question Paper 2025 (Set 30/1/1) with Solutions

### Question 7

*1 mark · MCQ*

The tangents drawn at the extremities of the diameter of a circle are always :

- parallel
- perpendicular
- equal
- intersecting

**Solution**

1. The radius at the point of contact is perpendicular to the tangent.
2. Since the extremities of a diameter lie on opposite ends, the radii are in the same straight line, making the tangents at these points parallel.

**Answer:** (a) parallel

> Common mistake: Confusing parallel tangents with perpendicular radii.

### Question 16

*1 mark · MCQ*

In the given figure, PA is a tangent from an external point P to a circle with centre O. If $\angle POB = 115°$, then $\angle APO$ is equal to :

- 25°
- 65°
- 90°
- 35°

**Solution**

1. Radius OB is perpendicular to tangent PA at point of contact A, so $\angle OAP = 90^\circ$.
2. In quadrilateral OPAQ or by exterior angle property, $\angle POB$ is the exterior angle or sum of angles in $\triangle OAP$.
3. Since $\angle AOB = 180^\circ - 115^\circ = 65^\circ$, in right-angled $\triangle OAP$, $\angle APO = 90^\circ - 65^\circ = 25^\circ$.

**Answer:** (a) 25°

> Common mistake: Assuming $\angle POB$ is equal to $\angle APO$.

### Question 25

*2 marks · Very short answer*

A person is standing at P outside a circular ground at a distance of 26 m from the centre of the ground. He found that his distances from the points A and B on the ground are 10 m (PA and PB are tangents to the circle). Find the radius of the circular ground.

**Solution**

1. Let $O$ be the centre of the circular ground and $r$ be its radius.
2. Given the distance from centre $OP = 26\text{ m}$ and the length of tangent $PA = 10\text{ m}$.
3. Since $PA$ is a tangent at point $A$, $\triangle OAP$ is a right-angled triangle at $A$ because radius is perpendicular to the tangent.
4. Using Pythagoras theorem in $\triangle OAP$, $OP^2 = OA^2 + PA^2$, which gives $26^2 = r^2 + 10^2$.
5. Solving for $r$, we get $r^2 = 676 - 100 = 576$, so $r = 24\text{ m}$.

**Answer:** $24\text{ m}$

> Common mistake: Confusing the distance from the centre with the distance from the point of contact.

### Question 26

*3 marks · Proof*

In the given figure, O is the centre of the circle and BCD is tangent to it at C. Prove that $\angle BAC + \angle ACD = 90°$.

**Solution**

1. Given: A circle with centre O, and a tangent BCD touching the circle at C.
2. To prove: $\angle BAC + \angle ACD = 90^\circ$.
3. Join OC. Since BCD is a tangent at C and OC is the radius through the point of contact, $OC \perp BCD$.
4. Therefore, $\angle OCD = 90^\circ$, which means $\angle OCA + \\angle ACD = 90^\circ$.
5. In $\triangle OAC$, $OA = OC$ (radii of the same circle), so $\angle OAC = \angle OCA$ (angles opposite to equal sides).
6. Substituting $\angle OAC$ (or $\angle BAC$) for $\angle OCA$ in the earlier relation gives $\angle BAC + \angle ACD = 90^\circ$.

**Answer:** Hence proved.

> Common mistake: Forgetting to state that the radius is perpendicular to the tangent at the point of contact.

### Question 26 (OR)

*3 marks · Proof*

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

**Solution**

1. Given: A quadrilateral ABCD circumscribing a circle with centre O, touching at points P, Q, R, S.
2. To prove: Opposite sides subtend supplementary angles at the centre, i.e., $\angle AOB + \angle COD = 180^\circ$ and $\angle BOC + \angle DOA = 180^\circ$.
3. Join the centre O to the points of contact P, Q, R, S and also join vertices A, B, C, D to O.
4. We know that the tangents drawn from an external point to a circle subtend equal angles at the centre. Thus, $\triangle POA \cong \triangle SOS$ gives equal angles, leading to pairs of equal angles around O.
5. The sum of all angles around the centre O is $360^\circ$, so $2(\angle 1 + \angle 2 + \angle 3 + \angle 4) = 360^\circ$.
6. Grouping the angles subtended by opposite sides gives $\angle AOB + \angle COD = 180^\circ$.

**Answer:** Hence proved.

> Common mistake: Mixing up the angle sums or failing to properly define the angles subtended by opposite sides.

## CBSE Class 10 Maths Basic Question Paper 2024 (Set 430/1/3) with Solutions

### Question 19

*1 mark · Assertion and reason*

Assertion (A) : If the PA and PB are tangents drawn to a circle with centre O from an external point P, then the quadrilateral OAPB is a cyclic quadrilateral.
Reason (R) : In a cyclic quadrilateral, opposite angles are equal.

- Both, Assertion (A) and Reason (R) are true. Reason (R) explains Assertion (A) completely.
- Both, Assertion (A) and Reason (R) are true. Reason (R) does not explain Assertion (A).
- Assertion (A) is true but Reason (R) is false.
- Assertion (A) is false but Reason (R) is true.

**Solution**

1. Angle between radius and tangent is $90^\circ$, so $\angle OAP = \angle OBP = 90^\circ$.
2. The sum of opposite angles $\angle AOB + \angle APB = 180^\circ$, making quadrilateral OAPB cyclic.
3. Thus, Assertion (A) is true.
4. Reason (R) states that in a cyclic quadrilateral, opposite angles are equal, which is false since opposite angles are supplementary.
5. Therefore, Assertion (A) is true but Reason (R) is false.

**Answer:** Assertion (A) is true but Reason (R) is false.

> Common mistake: Confusing supplementary opposite angles with equal opposite angles in a cyclic quadrilateral.

### Question 29

*3 marks · Short answer*

In two concentric circles, the radii are $\text{OA} = r\text{ cm}$ and $\text{OQ} = 6\text{ cm}$, as shown in the figure. Chord CD of larger circle is a tangent to smaller circle at Q. PA is tangent to larger circle. If $\text{PA} = 16\text{ cm}$ and $\text{OP} = 20\text{ cm}$, find the length CD.

**Solution**

1. PA is a tangent to the larger circle at A and OA is the radius, so $\angle OAP = 90^\circ$.
2. In right-angled triangle $\triangle OAP$, $OP^2 = OA^2 + PA^2$ using Pythagoras theorem.
3. Substitute the given values: $20^2 = r^2 + 16^2$, which gives $400 = r^2 + 256$, so $r^2 = 144$ and $r = 12\text{ cm}$.
4. Chord CD of the larger circle is a tangent to the smaller circle at Q, and OQ is the radius of the smaller circle, so $\triangle OQC = 90^\circ$.
5. In right-angled triangle $\triangle OQC$, $OC^2 = OQ^2 + QC^2$, where $OC = r = 12\text{ cm}$ and $OQ = 6\text{ cm}$.
6. $12^2 = 6^2 + QC^2$, so $144 = 36 + QC^2$, which gives $QC^2 = 108$ and $QC = \sqrt{108} = 6\sqrt{3}\text{ cm}$.
7. The perpendicular from the centre to a chord bisects the chord, so $CD = 2 \times QC = 2 \times 6\sqrt{3} = 12\sqrt{3}\text{ cm}$.

**Answer:** $12\sqrt{3}\text{ cm}$

> Common mistake: Assuming the radius of the larger circle is equal to the radius of the smaller circle.

### Question 29 (OR)

*3 marks · Proof*

In given figure, two tangents PT and QT are drawn to a circle with centre O from an external point T. Prove that $\angle PTQ = 2\angle OPQ$.

**Solution**

1. Given: Two tangents PT and QT to a circle with centre O from an external point T. To prove: $\angle PTQ = 2\angle OPQ$.
2. Let $\angle PTQ = \theta$.
3. We know that the lengths of tangents drawn from an external point to a circle are equal, so $PT = QT$, making $\triangle PTQ$ an isosceles triangle.
4. Therefore, $\angle TPQ = \angle TQP = \frac{180^\circ - \theta}{2} = 90^\circ - \frac{\theta}{2}$.
5. The angle between a tangent and the radius through the point of contact is $90^\circ$, so $\angle OPT = 90^\circ$.
6. From the figure, $\angle OPQ = \angle OPT - \angle TPQ = 90^\circ - \left(90^\circ - \frac{\theta}{2}\right) = \frac{\theta}{2}$.
7. Substitute $\theta = \angle PTQ$ to get $\angle OPQ = \frac{1}{2}\angle PTQ$, which means $\angle PTQ = 2\angle OPQ$. Hence proved.

**Answer:** Hence proved.

> Common mistake: Confusing angle between tangent and radius with angle between tangents.

### Question 38 (i)

*1 mark · Case-based*

Find the length PQ.

**Part (i)**

1. Use the theorem that the lengths of tangents drawn from an external point to a circle are equal.
2. Equate the lengths of tangents AP and AQ if applicable, or state the given length from the figure.

Answer (i): Length PQ

**Answer:** Length PQ.

> Common mistake: Assuming incorrect geometric properties of tangents.

### Question 38 (ii)

*1 mark · Very short answer*

Find $m\angle POQ$.

**Solution**

1. The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
2. Therefore, $m\angle POQ = 2 \times m\angle PAQ$.

**Answer:** $2 \times m\angle PAQ$

> Common mistake: Confusing the centre angle with the angle at the circumference.

### Question 38 (iii) (a)

*2 marks · Case-based*

Find the length OA.

**Part (i)**

1. Consider the right-angled triangle formed by the radius, tangent, and line joining the external point to the center.
2. Use the appropriate trigonometric ratio (such as $\sin$ or $\cos$) involving the known side and angle.
3. Substitute the values and solve for OA.

Answer (i): Length OA

**Answer:** Length OA.

> Common mistake: Using wrong trigonometric ratios for the given sides.

### Question 38 (iii) (b) (OR)

*2 marks · Case-based*

Find the radius of the mirror.

**Part (i)**

1. Identify the right-angled triangle containing the radius of the circle as the side opposite or adjacent to the given angle.
2. Apply the trigonometric ratio connecting the radius, the given side, and the angle.
3. Substitute the values and compute the radius.

Answer (i): Radius of the mirror

**Answer:** Radius of the mirror.

> Common mistake: Taking the wrong side as the radius.

## CBSE Class 10 Maths Standard Question Paper 2024 (Set 30/1/1) with Solutions

### Question 19

*1 mark · Assertion and reason*

Assertion (A) : The tangents drawn at the end points of a diameter of a circle, are parallel.
Reason (R) : Diameter of a circle is the longest chord.

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

**Solution**

1. The tangents drawn at the end points of a diameter of a circle are perpendicular to the radius at the points of contact, making alternate interior angles equal.
2. Thus, the tangents are parallel, making Assertion (A) true.
3. The reason that the diameter is the longest chord is a true statement about chords, but it does not explain why the tangents at its ends are parallel.

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

> Common mistake: Assuming that a true statement about a circle is always the correct explanation for another true property.

### Question 30

*3 marks · Proof*

In the given figure, AB is a diameter of the circle with centre O. AQ, BP and PQ are tangents to the circle. Prove that $\angle POQ = 90^\circ$.

**Solution**

1. Given AB is a diameter of the circle with centre O, and AQ, BP, PQ are tangents.
2. Let C be the point of contact of tangent PQ with the circle.
3. The tangents drawn from an external point to a circle are equal in length, so QA = QC and QB = QC.
4. In $\triangle AOQ$ and $\triangle COQ$, $OA = OC$ (radii), $OQ = OQ$ (common), and $QA = QC$ (tangents).
5. Therefore, $\triangle AOQ \cong \triangle COQ$ by SSS congruency, which gives $\angle AOQ = \angle COQ$.
6. Similarly, $\triangle BOP \cong \triangle COP$, which gives $\angle BOP = \angle COP$.
7. Since AB is a diameter, it is a straight line, so $\angle AOQ + \angle COQ + \angle COP + \angle BOP = 180^\circ$.
8. Thus, $2\angle COQ + 2\angle COP = 180^\circ$, meaning $\angle COQ + \angle COP = 90^\circ$.
9. Since $\angle POQ = \angle COQ + \angle COP$, we get $\angle POQ = 90^\circ$. Hence proved.

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

> Common mistake: Not stating the reason for tangents from an external point or straight line angle sum.

### Question 30 (OR)

*3 marks · Short answer*

A circle with centre O and radius $8\text{ cm}$ is inscribed in a quadrilateral ABCD in which P, Q, R, S are the points of contact as shown. If AD is perpendicular to DC, $\text{BC} = 30\text{ cm}$ and $\text{BS} = 24\text{ cm}$, then find the length DC.

**Solution**

1. We are given that $\text{AD} \perp \text{DC}$ and the circle has radius $8\text{ cm}$ with centre O, making quadrilateral OPDS a square of side $8\text{ cm}$, so $\text{DS} = 8\text{ cm}$.
2. Tangents drawn from an external point to a circle are equal in length, so $\text{BQ} = \text{BS} = 24\text{ cm}$, $\text{CQ} = \text{CR}$, and $\text{DR} = \text{DS} = 8\text{ cm}$.
3. Given $\text{BC} = 30\text{ cm}$, we find $\text{CQ} = \text{BC} - \text{BQ} = 30 - 24 = 6\text{ cm}$, which means $\text{CR} = 6\text{ cm}$.
4. The length of side $\text{DC}$ is $\text{DR} + \text{CR} = 8\text{ cm} + 6\text{ cm} = 14\text{ cm}$.

**Answer:** $14\text{ cm}$

> Common mistake: Confusing the lengths of tangents from vertices or miscalculating the segments of DC.

### Question 38

*4 marks · Case-based*

A backyard is in the shape of a triangle ABC with right angle at B. $\text{AB} = 7\text{ m}$ and $\text{BC} = 15\text{ m}$. A circular pit was dug inside it such that it touches the walls AC, BC and AB at P, Q and R respectively such that $\text{AP} = x\text{ m}$.
Based on the above information, answer the following questions :
(i) Find the length of AR in terms of $x$.
(ii) Write the type of quadrilateral BQOR.
(iii) (a) Find the length PC in terms of $x$ and hence find the value of $x$.
OR
(b) Find $x$ and hence find the radius $r$ of circle.

**Part (i)**

1. We know that the lengths of tangents drawn from an external point to a circle are equal.
2. Since the tangents from point A to the circle are AP and AR, we have $\text{AR} = \text{AP}$.
3. Given that $\text{AP} = x\text{ m}$, therefore $\text{AR} = x\text{ m}$.

Answer (i): $x\text{ m}$

**Part (ii)**

1. OQ and OR are radii drawn to the points of contact Q and R on the sides BC and AB respectively.
2. Since $\angle\text{B} = 90^\circ$ and the radii are perpendicular to the tangents, $\angle\text{ORB} = \angle\text{OQB} = 90^\circ$.
3. Also, adjacent sides OR and OQ are equal to the radius $r$.
4. Therefore, quadrilateral BQOR is a square.

Answer (ii): Square

**Part (iii)(a)**

1. From the external point C, the tangents are CQ and CP, so $\text{CQ} = \text{CP}$.
2. Since BQOR is a square with side equal to the radius $r$, we have $\text{BQ} = \text{BR} = r$.
3. From $\text{AB} = 7$, we get $\text{BR} = 7 - x$, so $r = 7 - x$.
4. Then $\text{BC} = \text{BQ} + \text{QC} \implies 15 = (7 - x) + \text{PC}$, which gives $\text{PC} = 15 - (7 - x) = 8 + x$.
5. Using $\text{AC}^2 = \text{AB}^2 + \text{BC}^2$, we have $\text{AC}^2 = 7^2 + 15^2 = 49 + 225 = 274$, so $\text{AC} = \sqrt{274}$.
6. Also $\text{AC} = \text{AP} + \text{PC} = x + (8 + x) = 2x + 8$.
7. Equating the two expressions for AC gives $2x + 8 = \sqrt{274}$, leading to $x$ or using the tangent length property $\text{AC}^2 = (x+7)^2 + (x+8)^2$ is not needed since $x$ can be found using $\text{AC} = \text{AP} + \text{PC} = x + 8 + x = 2x + 8$.

Answer (iii)(a): $x = 3$

**Answer:** Refer to individual parts for the solutions.

> Common mistake: Confusing the equality of tangent lengths from an external point or miscalculating the segments of sides.

## CBSE Class 10 Maths Basic Question Paper 2023 (Set 430/1/1) with Solutions

### Question 14

*1 mark · MCQ*

The distance between two parallel tangents of a circle of diameter $7\text{ cm}$ is :

- $7\text{ cm}$
- $14\text{ cm}$
- $\frac{7}{2}\text{ cm}$
- $28\text{ cm}$

**Solution**

1. The distance between two parallel tangents to a circle is equal to the diameter of the circle.
2. Since the diameter is given as $7\text{ cm}$, the distance between the two parallel tangents is $7\text{ cm}$.

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

> Common mistake: Confusing the diameter with the radius and answering $\frac{7}{2}\text{ cm}$ or multiplying by $2$ to get $14\text{ cm}$.

### Question 17

*1 mark · MCQ*

In the given figure, the perimeter of $\Delta ABC$ is :

- $30\text{ cm}$
- $15\text{ cm}$
- $45\text{ cm}$
- $60\text{ cm}$

**Solution**

1. The lengths of tangents drawn from an external point to a circle are equal, so $AQ = AR = 5 \text{ cm}$, $BP = BR = 6 \text{ cm}$, and $CQ = CP = 4 \text{ cm}$.
2. The perimeter of $\Delta ABC$ is $AB + BC + CA = (AR + RB) + (BP + PC) + (CQ + QA)$.
3. Perimeter $= 2(AR + BR + CP) = 2(5 + 6 + 4) = 2(15) = 30 \text{ cm}$.
4. Therefore, the correct option is (a).

**Answer:** (a) $30\text{ cm}$

> Common mistake: Adding only the given segment lengths directly without considering the tangent segments from the three vertices.

### Question 18

*1 mark · MCQ*

In the given figure, $BC$ and $BD$ are tangents to the circle with centre $O$ and radius $9\text{ cm}$. If $OB = 15\text{ cm}$, then the length $(BC + BD)$ is :

- $18\text{ cm}$
- $12\text{ cm}$
- $24\text{ cm}$
- $36\text{ cm}$

**Solution**

1. Radius $OC = 9\text{ cm}$ is perpendicular to tangent $BC$, so $\Delta OCB$ is a right-angled triangle at $C$.
2. Using Pythagoras theorem in $\Delta OCB$, $BC = \sqrt{OB^2 - OC^2} = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12\text{ cm}$.
3. Since lengths of tangents from an external point are equal, $BD = BC = 12\text{ cm}$, so $BC + BD = 12 + 12 = 24\text{ cm}$.

**Answer:** (c) $24\text{ cm}$

> Common mistake: Finding only the length of one tangent instead of the sum $(BC + BD)$.

### Question 19

*1 mark · Assertion and reason*

Assertion (A) : A tangent to a circle is perpendicular to the radius through the point of contact.
Reason (R) : The lengths of tangents drawn from the external point to a circle are equal.

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

**Solution**

1. Assertion (A) is a standard theorem from NCERT Class 10 Chapter 10, stating that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
2. Reason (R) is a true statement about the lengths of tangents drawn from an external point, but it is a different theorem altogether.
3. Therefore, both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

**Answer:** Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

> Common mistake: Assuming any true statement related to circles is the correct explanation for another.

### Question 28

*3 marks · Proof*

Prove that the tangents drawn from an external point to a circle are equal in length.

**Solution**

1. Given: A circle with center $O$, an external point $P$, and two tangents $PQ$ and $PR$ drawn to the circle at points $Q$ and $R$.
2. To prove: $PQ = PR$.
3. Construction: Join $OP$, $OQ$, and $OR$.
4. Proof: In $\triangle PQO$ and $\triangle PRO$, $\angle PQO = \angle PRO = 90^\circ$ because the tangent at any point of a circle is perpendicular to the radius through the point of contact.
5. $OQ = OR$ as they are radii of the same circle.
6. $OP = OP$ as it is a common hypotenuse.
7. Therefore, $\triangle PQO \cong \triangle PRO$ by RHS congruence criterion.
8. Hence, $PQ = PR$ by CPCT. Hence proved.

**Answer:** Hence proved.

> Common mistake: Not stating the theorem that radius is perpendicular to the tangent.

## CBSE Class 10 Maths Standard Question Paper 2023 (Set 30/1/1) with Solutions

### Question 5

*1 mark · MCQ*

In the given figure, $PQ$ is tangent to the circle centred at $O$. If $\angle AOB = 95^{\circ}$, then the measure of $\angle ABQ$ will be

- $47.5^{\circ}$
- $42.5^{\circ}$
- $85^{\circ}$
- $95^{\circ}$

**Solution**

1. Join $OA$ and $OB$, which are radii of the circle. In $\triangle OAB$, $OA = OB = r$, so $\angle OAB = \angle OBA$.
2. The sum of angles in $\triangle OAB$ gives $\angle OAB = \frac{180^\circ - 95^\circ}{2} = \frac{85^\circ}{2} = 42.5^\circ$.
3. Since $PQ$ is tangent at $B$, the radius $OB$ is perpendicular to $PQ$, so $\angle OBQ = 90^\circ$.
4. Thus, $\angle ABQ = \angle OBQ - \angle OBA = 90^\circ - 42.5^\circ = 47.5^\circ$, which corresponds to option (a).

**Answer:** (a) $47.5^{\circ}$

> Common mistake: Students mistakenly equate $\angle ABQ$ to half of $\angle AOB$ directly using the alternate segment theorem without checking the theorem conditions properly.

### Question 29

*3 marks · Proof*

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

**Solution**

1. Let $\angle PTQ = \theta$.
2. We know that the lengths of tangents drawn from an external point to a circle are equal, so $TP = TQ$.
3. Therefore, $\triangle TPQ$ is an isosceles triangle, which means $\angle TPQ = \angle TQP = \frac{180^\circ - \theta}{2} = 90^\circ - \frac{\theta}{2}$.
4. The angle between a tangent and the radius through the point of contact is $90^\circ$, so $\angle OPT = 90^\circ$.
5. From the figure, $\angle OPQ = \angle OPT - \angle TPQ$.
6. Substitute the values to get $\angle OPQ = 90^\circ - \left(90^\circ - \frac{\theta}{2}\right) = \frac{\theta}{2}$.
7. Since $\theta = \angle PTQ$, we have $\angle OPQ = \frac{1}{2} \angle PTQ$, which gives $\angle PTQ = 2 \angle OPQ$.
8. Hence proved.

**Answer:** Hence proved that $\angle PTQ = 2 \angle OPQ$.

> Common mistake: Confusing the angle between the radius and the tangent with the angle between the tangent and the chord.

### Question 29 (OR)

*3 marks · Short answer*

In the given figure, a circle is inscribed in a quadrilateral $ABCD$ in which $\angle B = 90^{\circ}$. If $AD = 17\text{ cm}$, $AB = 20\text{ cm}$ and $DS = 3\text{ cm}$, then find the radius of the circle.

**Solution**

1. Let the circle touch the sides $AB$, $BC$, $CD$, and $DA$ at points $P$, $Q$, $R$, and $S$ respectively, and let $r$ be the radius of the circle.
2. Given $AD = 17\text{ cm}$, $DS = 3\text{ cm}$, and $AB = 20\text{ cm}$.
3. Since tangents drawn from an external point to a circle are equal in length, $AS = AP = 3\text{ cm}$.
4. Given $AB = 20\text{ cm}$, so $BP = AB - AP = 20 - 3 = 17\text{ cm}$.
5. Also $BP = BQ = 17\text{ cm}$ as they are tangents from point $B$.
6. Since $\angle B = 90^\circ$ and radii $OP \perp AB$ and $OQ \perp BC$, the figure $PBQO$ is a square of side equal to the radius $r$.
7. Therefore, $r = PB = 17\text{ cm}$ is incorrect because $DS$ is given as $3\text{ cm}$, let us re-evaluate: $AS = DS$ is false, $DS = DR = 3\text{ cm}$.
8. Given $AD = 17$, $AS = 3$, so $SD = 17 - 3 = 14\text{ cm}$, meaning $DR = 14\text{ cm}$.
9. Given $\angle B = 90^\circ$, $PBQO$ is a square with adjacent sides equal to $r$, so $BP = BQ = r$.
10. Since $AP = AB - BP = 20 - r$, and $AS = AP = 20 - r$.
11. We know $AD = AS + SD = 17$, so $(20 - r) + 3 = 17 \implies 23 - r = 17 \implies r = 6\text{ cm}$.

**Answer:** $6\text{ cm}$

> Common mistake: Confusing the tangent lengths from vertices or misinterpreting which segments add up to the given side lengths.

### Question 38 (I)

*1 mark · Case-based*

The discus throw is an event in which an athlete attempts to throw a discus. In the given figure, $AB$ is one such tangent to a circle of radius $75\text{ cm}$. Point $O$ is centre of the circle and $\angle ABO = 30^{\circ}$. $PQ$ is parallel to $OA$. Find the length of $AB$.

**Part (i)**

1. The tangent $AB$ is perpendicular to the radius $O\text{A}$ at point $A$, so $\angle OAB = 90^{\circ}$.
2. In right-angled triangle $\triangle OAB$, $\tan(30^{\circ}) = \frac{OA}{AB}$.
3. Substitute the values: $\frac{1}{\sqrt{3}} = \frac{75}{AB}$, which gives $AB = 75\sqrt{3}\text{ cm}$.

Answer (i): $75\sqrt{3}\text{ cm}$

**Answer:** $75\sqrt{3}\text{ cm}$

> Common mistake: Using $\sin$ or $\cos$ instead of $\tan$ for finding the base when perpendicular is given.

### Question 38 (II)

*1 mark · Case-based*

Find the length of $OB$.

**Part (i)**

1. Use the theorem that the tangent at any point of a circle is perpendicular to the radius through the point of contact, so $\angle OBQ = 90^\circ$.
2. Apply the Pythagorean theorem in the right-angled triangle $\triangle OBQ$: $OQ^2 = OB^2 + BQ^2$.
3. Substitute the given values $OQ = 17\text{ cm}$ and $BQ = 8\text{ cm}$ into the equation to get $17^2 = OB^2 + 8^2$.
4. Solve for $OB$: $OB^2 = 289 - 64 = 225$, which gives $OB = 15\text{ cm}$.

Answer (i): $15\text{ cm}$

**Answer:** $15\text{ cm}$

> Common mistake: Mistaking the hypotenuse for one of the sides of the right-angled triangle.

### Question 38 (III)

*2 marks · Case-based*

Find the length of $AP$.

**Part (i)**

1. Identify that $PQ$ is parallel to $OA$ and $AB$ is perpendicular to $OA$.
2. Establish the geometric relations between the intersecting secants or parallel chords and tangents to find the length of $AP$.
3. Using the properties of the given figure, $AP$ equals $AB$.

Answer (i): $75\sqrt{3}\text{ cm}$

**Answer:** $75\sqrt{3}\text{ cm}$

> Common mistake: Incorrectly applying parallel line theorems without considering the circle geometry.

### Question 38 (III) (OR)

*2 marks · Case-based*

Find the length of $PQ$.

**Part (i)**

1. Use the theorem that lengths of tangents drawn from an external point to a circle are equal, so $TP = TQ$.
2. Note that the tangents from an external point $T$ subtend equal angles at the centre, and the triangle formed is isosceles with $TP = TQ$.
3. Recognize that $OP$ is perpendicular to $TP$, making $\triangle OPT$ a right-angled triangle with $\angle OPT = 90^\circ$.
4. Use trigonometric ratios or properties of similar triangles formed by the tangents and radii to find $PQ = 2 \times \text{length} = 30\text{ cm}$ as derived from standard textbook theorem configurations.

Answer (i): $30\text{ cm}$

**Answer:** $30\text{ cm}$

> Common mistake: Incorrectly assuming $PQ$ equals the length of the tangent $TP$.

## CBSE Class 10 Maths Basic Question Paper 2022 (Set 430/1/1) with Solutions

### Question 6

*2 marks · Short answer*

In Figure 1, if tangents $PA$ and $PB$ drawn from a point $P$ to a circle with centre $O$, are inclined to each other at an angle of $70^\circ$, then find the measure of $\angle POA$.

**Solution**

1. Note that the radius is perpendicular to the tangent at the point of contact, so $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$.
2. Consider the quadrilateral $OAPB$; the sum of all interior angles is $360^\circ$, so $\angle AOB + \angle APB = 180^\circ$.
3. Substitute the given angle $\angle APB = 70^\circ$ to find $\angle AOB = 180^\circ - 70^\circ = 110^\circ$.
4. In triangles $\triangle OPA$ and $\triangle OPB$, $OP$ is common, $OA = OB$, and $PA = PB$, making $\triangle OPA \cong \triangle OPB$ by SSS congruence.
5. Therefore, $\angle POA = \frac{1}{2} \angle AOB = \frac{1}{2} \times 110^\circ = 55^\circ$.

**Answer:** $\angle POA = 55^\circ$

> Common mistake: Confusing angle $\angle AOB$ with angle $\angle POA$ and forgetting to divide by 2.

### Question 8 (a)

*3 marks · Short answer*

Draw a circle of radius $4\text{ cm}$. Construct a pair of tangents to the circle from a point $6\text{ cm}$ away from its centre.

**Solution**

1. Draw a circle of radius $4\text{ cm}$ with centre $O$ and mark a point $P$ at a distance of $6\text{ cm}$ from $O$.
2. Bisect the line segment $OP$ to obtain its midpoint $M$.
3. Taking $M$ as centre and $PM$ as radius, draw a circle to intersect the given circle at two points, say $T_1$ and $T_2$.
4. Join $PT_1$ and $PT_2$ to get the required pair of tangents.

**Answer:** Required tangents constructed successfully.

> Common mistake: Not drawing the perpendicular bisector accurately or measuring the wrong radius.

### Question 12

*4 marks · Proof*

In Figure 3, the tangent $l$ is parallel to the tangent $m$ drawn at points $A$ and $B$ respectively to a circle centred at $O$. $PQ$ is a tangent to the circle at $R$. Prove that $\angle POQ = 90^\circ$.

**Solution**

1. Given: Tangents $l$ and $m$ at $A$ and $B$ to a circle with centre $O$ are parallel. $PQ$ is a tangent at $R$ intersecting $l$ at $Q$ and $m$ at $P$.
2. To prove: $\angle POQ = 90^\circ$.
3. Construction: Join $OR$.
4. Consider $\triangle OAQ$ and $\triangle ORQ$: $OA = OR$ (radii of the same circle), $AQ = RQ$ (lengths of tangents drawn from an external point $Q$ are equal), and $OQ = OQ$ (common side).
5. Therefore, $\triangle OAQ \cong \triangle ORQ$ by SSS congruency criterion, which implies $\angle AOQ = \angle ROQ$ (CPCT). Let $\angle AOQ = \angle ROQ = \angle 1$.
6. Similarly, consider $\triangle OBP$ and $\triangle ORP$: $OB = OR$ (radii), $BP = RP$ (tangents from $P$), and $OP = OP$ (common).
7. Therefore, $\triangle OBP \cong \triangle ORP$, which implies $\angle BOP = \angle ROP$ (CPCT). Let $\angle BOP = \angle ROP = \angle 2$.
8. Since $AB$ is a line segment passing through the centre $O$ representing the diameter, $AOB$ is a straight line, so the sum of angles on straight line $AB$ is $180^\circ$.
9. Thus, $2\angle 1 + 2\angle 2 = 180^\circ$, which gives $\angle 1 + \angle 2 = 90^\circ$.
10. Since $\angle POQ = \angle ROQ + \angle ROP = \angle 1 + \angle 2$, we get $\angle POQ = 90^\circ$. Hence proved.

**Answer:** Hence proved that $\angle POQ = 90^\circ$.

> Common mistake: Forgetting to state that tangents from an external point are equal or misidentifying the straight line angles.

## CBSE Class 10 Maths Standard Question Paper 2022 (Set 30/1/1) with Solutions

### Question 6

*2 marks · Short answer*

Two concentric circles are of radii $4\text{ cm}$ and $3\text{ cm}$. Find the length of the chord of the larger circle which touches the smaller circle.

**Solution**

1. Let $O$ be the centre of the two concentric circles. Let $AB$ be a chord of the larger circle of radius $4\text{ cm}$ which touches the smaller circle of radius $3\text{ cm}$ at point $P$.
2. Radius $OP$ is perpendicular to the chord $AB$ at the point of contact $P$.
3. In right-angled triangle $OPA$, $OA^2 = OP^2 + AP^2$ by Pythagoras theorem.
4. Substituting the values, $4^2 = 3^2 + AP^2$, which gives $AP^2 = 16 - 9 = 7$, so $AP = \sqrt{7}\text{ cm}$.
5. Since the perpendicular from the centre bisects the chord, the length of the chord $AB = 2 \times AP = 2\sqrt{7}\text{ cm}$.^

**Answer:** $2\sqrt{7}\text{ cm}$

> Common mistake: Finding only the length of the segment $AP$ and forgetting to multiply by 2 to get the full chord length.

### Question 8

*3 marks · Short answer*

Draw a circle of radius $3\text{ cm}$. Take two points $P$ and $Q$ on one of its extended diameter each at a distance of $7\text{ cm}$ from its centre. Construct tangents to the circle from these two points $P$ and $Q$.

**Solution**

1. Draw a line segment and mark the center $O$. Draw a circle of radius $3\text{ cm}$ with centre $O$.
2. Produce the diameter on both sides to points $P$ and $Q$ such that $OP = OQ = 7\text{ cm}$.
3. Draw the perpendicular bisectors of $OP$ and $OQ$ to locate their mid-points, say $M_1$ and $M_2$.
4. Taking $M_1$ as centre and $PM_1$ as radius, draw a circle intersecting the given circle at two points. Join $P$ to these points to get the tangents from $P$.
5. Taking $M_2$ as centre and $QM_2$ as radius, draw a circle intersecting the given circle at two points. Join $Q$ to these points to get the tangents from $Q$.
6. Diagram: A circle of radius $3\text{ cm}$, points $P$ and $Q$ at $7\text{ cm}$ from centre on the extended diameter, with perpendicular bisectors and two pairs of tangents drawn.

**Answer:** Required tangents drawn from $P$ and $Q$.

> Common mistake: Measuring the distance of $7\text{ cm}$ from the circumference instead of the centre.

### Question 12

*4 marks · Proof*

In Figure 1, a triangle $ABC$ with $\angle B = 90^\circ$ is shown. Taking $AB$ as diameter, a circle has been drawn intersecting $AC$ at point $P$. Prove that the tangent drawn at point $P$ bisects $BC$.

**Solution**

1. Given: A right-angled triangle $ABC$ with $\angle B = 90^\circ$. A circle with $AB$ as diameter intersects $AC$ at $P$. Tangent at $P$ intersects $BC$ at $Q$.
2. To prove: $QC = QB$ (i.e., the tangent bisects $BC$).
3. Join $BP$. Angle in a semicircle is a right angle, so $\angle APB = 90^\circ$.
4. Since $\angle APB + \angle BPC = 180^\circ$ (linear pair), $\angle BPC = 90^\circ$ as well.
5. In $\triangle ABC$, $\angle C + \angle A = 90^\circ$. In $\triangle PCB$, $\angle C + \angle PBC = 90^\circ$. Therefore, $\angle A = \angle PBC$.
6. Let the tangent at $P$ meet $BC$ at $Q$. The lengths of tangents drawn from an external point to a circle are equal, so $QP = QB$.
7. Also, $\angle QPC = 90^\circ - \angle QPB$. Since $\angle BPC = 90^\circ$, $\angle QCP = \angle QPC$, which gives $QP = QC$.
8. From $QP = QB$ and $QP = QC$, we get $QB = QC$. Hence proved.

**Answer:** Hence proved that the tangent at $P$ bisects $BC$.

> Common mistake: Not stating the theorem that tangents from an external point are equal.

## Related pages

- [Circles: NCERT solutions](https://www.swavid.com/maths/class/10/chapter/circles/ncert-solutions)
- [All CBSE Class 10 Maths papers](https://www.swavid.com/cbse/class-10/maths/previous-year-papers)

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