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CBSE · Class 10 · Mathematics

Circles

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
From a point P, 26 cm from the centre O of a circle of radius 24 cm, a tangent PQ is drawn. The length of PQ is:
  1. 50 cm
  2. 2 cm
  3. 11 cm
  4. 10 cm
2.
The tangent from a point P to a circle of radius 24 cm is 18 cm long. The distance of P from the centre is:
  1. 42 cm
  2. 6 cm
  3. 30 cm
  4. 31 cm
3.
PA and PB are tangents from P to a circle with centre O. If ∠APB = 115°, then ∠AOB is:
  1. 245°
  2. 115°
  3. 32.5°
  4. 65°
4.
PA is a tangent at A to a circle with centre O. If ∠AOP = 72°, then ∠APO is:
  1. 108°
  2. 162°
  3. 72°
  4. 18°
5.
A tangent to a circle meets it at:
  1. exactly one point
  2. two points
  3. no point
  4. three points
6.
Two concentric circles have radii 50 cm and 14 cm. The chord of the larger circle that touches the smaller circle is:
  1. 72 cm
  2. 96 cm
  3. 48 cm
  4. 36 cm
7.
A quadrilateral ABCD is drawn to circumscribe a circle. If AB = 10 cm, BC = 9 cm and CD = 12 cm, then DA is:
  1. 22 cm
  2. 10 cm
  3. 13 cm
  4. 2 cm
8.
Assertion (A): The tangent to a circle of radius 6 cm from a point 10 cm from its centre is 8 cm long.
Reason (R): The tangent at any point of a circle is perpendicular to the radius through the point of contact.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
Find the length of the tangent drawn to a circle of radius 105 cm from a point 111 cm from its centre.
10.
Tangents PA and PB are drawn from an external point P to a circle with centre O. If ∠AOB = 155°, find ∠APB.
11.
PA and PB are tangents from P to a circle. If ∠APB = 48°, find ∠PAB.
12.
A circle is inscribed in △ABC, touching AB, BC and CA at P, Q and R. If AB = 18 cm, BC = 21 cm and CA = 19 cm, find AP, BQ and CR.
13.
A chord of the larger of two concentric circles is 24 cm long and touches the smaller circle, whose radius is 16 cm. Find the radius of the larger circle.
14.
ABCD circumscribes a circle. AB = 14 cm, BC = 19 cm and CD = 20 cm. Find DA and the perimeter of ABCD.
15.
The tangent from a point 45 cm from the centre of a circle is 27 cm long. Find the radius of the circle.
16.
From a point P, 50 cm from the centre O of a circle of radius 40 cm, tangents PA and PB are drawn. Find the length of AB.
Section C — Word problems4 × 4 = 16 marks
17.
A circular park has a radius of 12 m. Isha stands 15 m from its centre and walks in a straight line that just touches the edge of the park. How far does Isha walk to reach the edge?
18.
A circular pond touches all three sides of a triangular plot ABC. AB = 35 m, BC = 51 m and CA = 32 m. How far from corner A are the two points where the pond touches the sides?
19.
Two tangents drawn from a point P to a circle of radius 17 cm are inclined to each other at 90°. Find the distance OP from the centre and the length of each tangent.
20.
A quadrilateral field ABCD has a circular pond that touches all four sides. AB = 29 m, BC = 32 m and CD = 20 m. Find DA, and the cost of fencing the field at Rs 45 per metre.
AmbifulWorksheets
Practice that never repeats
CBSE · Class 10 · Mathematics

Answer key

Circles · AW-10M-CIR-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) 10 cmOQ ⟂ PQ, so PQ = √(262 − 242) = 10
2(c) 30 cmOP = √(242 + 182) = 30
3(d) 65°∠OAP = ∠OBP = 90°, so ∠AOB = 360° − 180° − 115° = 65°
4(d) 18°∠OAP = 90°, so ∠APO = 90° − 72° = 18°
5(a) exactly one point—
6(b) 96 cmthe radius to the point of contact bisects the chord: half-chord = √(502 − 142) = 48
7(c) 13 cmAB + CD = BC + DA: DA = 10 + 12 − 9 = 13
8(a) Both A and R are true, and R is the correct explanation of A.A is true: √(102 − 62) = 8. R is true and explains A.
936 cm√(1112 − 1052) = √1296 = 36
1025°OAPB has two right angles, so ∠APB = 180° − 155° = 25°
1166°PA = PB, so △PAB is isosceles: ∠PAB = (180° − 48°)/2
12AP = 8 cm, BQ = 10 cm, CR = 11 cmtangents from a point are equal: AP = AR, BP = BQ, CQ = CR; AP = (AB + CA − BC)/2 = 8
1320 cmhalf-chord 12; R = √(162 + 122) = 20
14DA = 15 cm; perimeter 68 cmAB + CD = BC + DA; perimeter = 2(AB + CD)
1536 cmr = √(452 − 272) = 36
1648 cmPA = 30; OP bisects AB at right angles, and (1/2) × OA × PA = (1/2) × OP × (AB/2), so AB = 2 × 40 × 30/50
179 mthe path is a tangent, at right angles to the radius: √(152 − 122) = 9
188 m (on both AB and AC)the tangents from A are equal; AP = (AB + AC − BC)/2 = (35 + 32 − 51)/2 = 8
19OP = 17√2 cm; each tangent 17 cmOP bisects the angle, so ∠APO = 45°; sin 45° = 17/OP and tan 45° = 17/PA
20DA = 17 m; Rs 4,410AB + CD = BC + DA; perimeter = 98 m