CBSE Class 10 · Chapter 10 · Geometry · 2026-27

Circles Class 10: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceTangent perpendicular to the radius

A point $P$ is $13$ cm from the centre of a circle of radius $5$ cm. The length of the tangent from $P$ to the circle is

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Answer: (a)

12

Why: The radius to the point of contact is perpendicular to the tangent: 13² − 5² = 144.

$△ O T P$ is right-angled at the point of contact $T$ : $P T = \sqrt{13^{2} - 5^{2}} = \sqrt{144} = 12$ cm.

Q2

·1 mark·Multiple choiceLengths of tangents from an external point

From a point $P$ outside a circle, tangents are drawn to the circle. Which statement is correct?

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Answer: (b) Exactly two tangents can be drawn, and they are equal in length

Why: From an external point there are exactly two tangents, and their lengths are equal.

From a point outside a circle exactly two tangents can be drawn (and they are equal in length).

Q3

·1 mark·Multiple choiceTangent perpendicular to the radius

A line that meets a circle at exactly one point is called

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Answer: (b) a tangent

Why: A secant meets the circle twice; a tangent touches it once.

A secant cuts the circle at two points; a tangent meets it at exactly one point, the point of contact.

Q4

·1 mark·Multiple choiceLengths of tangents from an external point

$P A$ and $P B$ are tangents from $P$ to a circle with centre $O$ . If $∠ A P B = 70^{\circ}$ , then $∠ A O B$ is

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Answer: (a)

110^{\circ}

Why: In quadrilateral OAPB the angles at A and B are 90°, so ∠AOB = 180° − ∠APB.

$∠ O A P = ∠ O B P = 90^{\circ}$ . Angle sum of $O A P B$ : $∠ A O B = 360^{\circ} - 90^{\circ} - 90^{\circ} - 70^{\circ} = 110^{\circ}$ .

Q5

·1 mark·Multiple choiceLengths of tangents from an external point

A quadrilateral $A B C D$ is drawn to circumscribe a circle. If $A B = 6$ cm, $B C = 7$ cm and $C D = 5$ cm, then $D A$ is

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Answer: (c)

4

Why: Equal tangents give AB + CD = BC + DA.

Using equal tangent lengths from each vertex, $A B + C D = B C + D A$ : $6 + 5 = 7 + D A \Rightarrow D A = 4$ cm.

Q6

·1 mark·Multiple choiceTangent perpendicular to the radius

$P Q$ is a tangent at $Q$ to a circle with centre $O$ and radius $9$ cm. If $P Q = 12$ cm, then $O P$ is

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Answer: (b)

15

Why: OQ ⊥ PQ, so OP² = 9² + 12².

$O P = \sqrt{9^{2} + 12^{2}} = \sqrt{225} = 15$ cm.

Q7

·1 mark·Multiple choiceLengths of tangents from an external point

Tangents $P A$ and $P B$ from $P$ to a circle with centre $O$ make $∠ A P B = 60^{\circ}$ . If $P A = 6 \sqrt{3}$ cm, the radius of the circle is

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Answer: (d)

6

Why: OP bisects ∠APB, so in right △OAP, tan 30° = OA/PA.

$∠ A P O = 30^{\circ}$ and $∠ O A P = 90^{\circ}$ . $O A = P A \tan 30^{\circ} = 6 \sqrt{3} \times \frac{1}{\sqrt{3}} = 6$ cm.

Q8

·1 mark·Multiple choiceTangent perpendicular to the radius

The angle between a tangent to a circle and the radius drawn to the point of contact is

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Answer: (b)

90^{\circ}

Why: Theorem: the tangent is perpendicular to the radius at the point of contact.

The tangent at any point of a circle is perpendicular to the radius through the point of contact, so the angle is $90^{\circ}$ .

Q9

·1 mark·Multiple choiceTangent perpendicular to the radius

A chord of the larger of two concentric circles, of radii $25$ cm and $7$ cm, touches the smaller circle. Its length is

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Answer: (c)

48

Why: The chord is a tangent to the inner circle, so the radius 7 bisects it at right angles: half-chord = √(625 − 49).

The radius of the small circle meets the chord at right angles at its mid-point. Half-chord $= \sqrt{25^{2} - 7^{2}} = \sqrt{576} = 24$ cm, so the chord is $48$ cm.

Q10

·1 mark·Multiple choiceLengths of tangents from an external point

$P A$ and $P B$ are tangents from $P$ to a circle. If $∠ P A B = 65^{\circ}$ , then $∠ A P B$ is

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Answer: (b)

50^{\circ}

Why: PA = PB, so △PAB is isosceles with ∠PBA = 65° as well.

$P A = P B \Rightarrow ∠ P B A = ∠ P A B = 65^{\circ}$ . $∠ A P B = 180^{\circ} - 130^{\circ} = 50^{\circ}$ .

Q11

·1 mark·Multiple choiceTangent perpendicular to the radius

$P Q$ is a tangent at $Q$ to a circle with centre $O$ and radius $5$ cm. If $∠ P O Q = 60^{\circ}$ , then $P Q$ is

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Answer: (d)

5 \sqrt{3}

Why: △OQP is right-angled at Q: PQ = OQ tan 60°.

$∠ O Q P = 90^{\circ}$ , so $P Q = O Q \tan 60^{\circ} = 5 \sqrt{3}$ cm.

Q12

·1 mark·Multiple choiceLengths of tangents from an external point

A circle is inscribed in $△ A B C$ with $A B = 8$ cm, $B C = 10$ cm and $C A = 12$ cm. The length of the tangent from $A$ to the circle is

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Answer: (a)

5

Why: With tangent lengths x, y, z from A, B, C: x + y = 8, y + z = 10, z + x = 12.

Adding: $2 (x + y + z) = 30$ , so $x + y + z = 15$ and $x = 15 - (y + z) = 15 - 10 = 5$ cm.

Q13

·1 mark·Multiple choiceLengths of tangents from an external point

$P A$ and $P B$ are tangents from $P$ to a circle with centre $O$ . If $∠ P O A = 55^{\circ}$ , then the angle between the two tangents, $∠ A P B$ , is

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Answer: (c)

70^{\circ}

Why: In right △OAP, ∠APO = 90° − 55° = 35°, and OP bisects ∠APB.

$∠ O A P = 90^{\circ}$ , so $∠ A P O = 90^{\circ} - 55^{\circ} = 35^{\circ}$ . $O P$ bisects $∠ A P B$ (congruent $△ O A P$ , $△ O B P$ ), so $∠ A P B = 70^{\circ}$ .

Q14

·1 mark·Multiple choiceLengths of tangents from an external point

$P A$ and $P B$ are tangents from an external point $P$ to a circle, with $P A = (3 x - 2)$ cm and $P B = (x + 6)$ cm. The length $P A$ is

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Answer: (c)

10

Why: Tangents from an external point are equal: 3x − 2 = x + 6.

$3 x - 2 = x + 6 \Rightarrow x = 4$ , so $P A = 3 (4) - 2 = 10$ cm.

Q15

·1 mark·Multiple choiceLengths of tangents from an external point

A circle is inscribed in a triangle whose sides are $6$ cm, $8$ cm and $10$ cm (a right triangle). The radius of the circle is

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Answer: (d)

2

Why: At the right angle the two tangent lengths equal r, so (6 − r) + (8 − r) = 10.

The radii to the two legs and the legs form a square of side $r$ at the right angle. Tangent lengths: $6 - r$ and $8 - r$ along the legs, and these add to the hypotenuse: $14 - 2 r = 10 \Rightarrow r = 2$ cm.

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 10 case study questions by chapter.

Case study 1: Roundabout and roads (4 marks)

A circular roundabout has centre $O$ and radius $15$ m. A straight road touches the roundabout at $A$ , and a traffic post $P$ stands on this road $20$ m from $A$ . A second straight road from $P$ touches the roundabout at $B$ .

(i) What is the measure of

∠ O A P

? Give a reason. [1 mark]

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Answer:

90^{\circ}

90^{\circ}

: the tangent at a point is perpendicular to the radius through the point of contact. A1

(ii) Find the distance

O P

. [1 mark]

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Answer:

25

O P = \sqrt{15^{2} + 20^{2}} = \sqrt{625} = 25

m. A1

(iii) Find the length

P B

and the perimeter of the quadrilateral

O A P B

. [2 marks]

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Answer:

P B = 20

70

Tangents from

P

are equal:

P B = P A = 20

m A1. Perimeter

= 15 + 20 + 20 + 15 = 70

m. A1

OR Another post

Q

on the first road is

39

m from

O

. How far is

Q

from

A

? [2 marks]

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Answer:

36

∠ O A Q = 90^{\circ}

M1, so

Q A = \sqrt{39^{2} - 15^{2}} = \sqrt{1296} = 36

m. A1

Case study 2: Plate in a triangular tray (4 marks)

A circular plate fits exactly inside a triangular tray $A B C$ , touching the sides $B C$ , $C A$ and $A B$ at $D$ , $E$ and $F$ respectively. A student measures $A F = 4$ cm, $B D = 6$ cm and $C E = 8$ cm.

(i) Find

A E

, giving a reason. [1 mark]

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Answer:

4

A E = A F = 4

cm (tangents from an external point are equal). A1

(ii) Find the length of side

B C

. [1 mark]

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Answer:

14

B C = B D + D C = B D + C E = 6 + 8 = 14

cm. A1

(iii) Find the perimeter of the tray. [2 marks]

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Answer:

36

A B = 4 + 6 = 10

B C = 14

C A = 8 + 4 = 12

M1; perimeter

= 36

cm. A1

OR A second tray has

A B = 11

cm,

B C = 15

cm and

C A = 12

cm, with a plate touching all three sides. Find

A F

. [2 marks]

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Answer:

4

Let the tangent lengths from

A, B, C

x, y, z

x + y = 11

y + z = 15

z + x = 12

M1;

x + y + z = 19

, so

A F = x = 19 - 15 = 4

cm. A1

Next steps for Circles

This free set is separate from the chapter's question bank. On the Circles chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 54 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

Circles: important questions What Pro includes

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.