Class 10 Mathematics: Circles important questions

Circles is a chapter of Class 10 Mathematics for the CBSE. These 17 questions are marked Very Important, Important or Practice, and each answer has been checked by our team. Use them to revise, not as a prediction of what will come in the exam.

1. Practice · 1 marks

   1. If two tangents inclined at 60° are drawn to a circle of radius 3 cm, the length of each tangent is: (a) 3√3/2 cm (b) 3 cm (c) 6 cm (d) 3√3 cm

   (d) 3√3 cm The line from the external point to the centre bisects the 60° angle: tan 30° = 3/PA, so PA = 3√3 cm.

2. Important · 2 marks

   2. PA and PB are tangents to a circle with centre O, and ∠APB = 75°. Q is a point on the major arc. Find ∠AQB.

   ∠PAO = ∠PBO = 90°, so ∠AOB = 360° − 180° − 75° = 105°. ∠AQB = ½ ∠AOB = 52.5° (angle at the circle is half the angle at the centre).

3. Important · 3 marks

   3. Prove that a parallelogram circumscribing a circle is a rhombus.

   Let ABCD touch the circle at P, Q, R, S on AB, BC, CD, DA. Tangents from a point are equal: AP = AS, BP = BQ, CR = CQ, DR = DS. Adding: AB + CD = AD + BC. In a parallelogram AB = CD and AD = BC, so 2AB = 2AD and AB = AD. A parallelogram with equal adjacent sides is a rhombus.

4. Important · 3 marks

   4. XY and X′Y′ are two parallel tangents to a circle with centre O, and another tangent AB, touching the circle at C, meets XY at A and X′Y′ at B. Find ∠AOB.

   Let XY touch the circle at P and X′Y′ at Q. ΔOPA ≅ ΔOCA, so OA bisects ∠PAC; similarly OB bisects ∠QBC. XY ∥ X′Y′, so ∠PAB + ∠QBA = 180°, giving ∠OAB + ∠OBA = 90°. Hence ∠AOB = 180° − 90° = 90°.

5. Practice · 1 marks

   5. O is the centre of a circle, and chord PQ makes an angle of 50° with the tangent PR at the point of contact P. The angle subtended by the chord at the centre is: (a) 130° (b) 100° (c) 50° (d) 30°

   (b) 100° ∠OPQ = 90° − 50° = 40° = ∠OQP, so ∠POQ = 180° − 80° = 100°.

6. Practice · 1 marks

   6. A quadrilateral PQRS circumscribes a circle. If PQ = 12 cm, QR = 15 cm and RS = 14 cm, the length of SP is: (a) 15 cm (b) 14 cm (c) 12 cm (d) 11 cm

   (d) 11 cm PQ + RS = QR + SP, so 12 + 14 = 15 + SP and SP = 11 cm.

7. Important · 2 marks

   7. From an external point P, tangents PA and PB are drawn to a circle with centre O. A tangent at a point E on the circle meets PA and PB at C and D. If PA = 10 cm, find the perimeter of ΔPCD.

   Tangents from a point are equal: PA = PB, CA = CE, DB = DE. Perimeter = PC + CD + PD = PC + CE + ED + PD = PC + CA + DB + PD = PA + PB = 2PA = 20 cm.

8. Important · 3 marks

   8. PA and PB are tangents from an external point P to a circle with centre O. The chord AB makes an angle of 30° with the radius OA. If the chord is 6 cm long, find the length of the tangent PA and the radius OA.

   ∠OAP = 90°, so ∠PAB = 90° − 30° = 60°. PA = PB, so ∠PBA = 60° and ∠APB = 60°: ΔABP is equilateral, so PA = AB = 6 cm. In right ΔOAP, ∠OPA = 30° (OP bisects ∠APB), so tan 30° = OA/PA and OA = 6/√3 = 2√3 cm.

9. Important · 3 marks

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

   Let ∠TPQ = θ. ∠TPO = 90°, so ∠OPQ = 90° − θ. TP = TQ, so ∠TQP = θ and ∠PTQ = 180° − 2θ = 2(90° − θ) = 2∠OPQ.

10. Practice · 1 marks

    10. Which of the following statements is not true? (A) A number of secants can be drawn at any point on the circle. (B) Only one tangent can be drawn at any point on a circle. (C) A chord is a line segment joining two points on the circle. (D) From a point inside a circle only two tangents can be drawn.

    (D) is not true: no tangent can be drawn from a point inside a circle.

11. Practice · 1 marks

    11. PT is a tangent at a point P on a circle with centre O, and PQ is a chord. If ∠TPQ = 110°, then ∠POQ is: (A) 110° (B) 70° (C) 140° (D) 55°

    (C) 140° ∠OPT = 90°, so ∠OPQ = 110° − 90° = 20°. OP = OQ, so ∠OQP = 20° and ∠POQ = 180° − 40° = 140°.

12. Very Important · 5 marks

    12. Prove that the lengths of tangents drawn from an external point to a circle are equal. Using this, find BC in ΔABC, where a circle inscribed in ΔABC touches AB, BC and CA at R, P and Q respectively, AB = 10 cm, AQ = 7 cm and CQ = 5 cm.

    Proof: let PA and PB be tangents from P to a circle with centre O. Join OA, OB, OP. In right triangles OAP and OBP (∠OAP = ∠OBP = 90°), OA = OB (radii) and OP is common, so ΔOAP ≅ ΔOBP (RHS) and PA = PB. Application: AR = AQ = 7 cm, so BR = BP = 10 − 7 = 3 cm. CP = CQ = 5 cm. BC = BP + PC = 3 + 5 = 8 cm.

13. Practice · 1 marks

    13. A quadrilateral ABCD circumscribes a circle. If BC = 7 cm, CD = 4 cm and AD = 3 cm, then AB is: (A) 3 cm (B) 4 cm (C) 6 cm (D) 7 cm

    (C) 6 cm AB + CD = AD + BC, so AB + 4 = 3 + 7 and AB = 6 cm.

14. Practice · 1 marks

    14. The length of a tangent to a circle of radius 9 cm from a point 41 cm from the centre is: (A) 40 cm (B) 9 cm (C) 41 cm (D) 50 cm

    (A) 40 cm The tangent is perpendicular to the radius: TP² = 41² − 9² = 1681 − 81 = 1600, so TP = 40 cm.

15. Important · 2 marks

    15. A triangle ABC is drawn to circumscribe a circle of radius 4 cm, with the point of contact D on BC such that BD = 10 cm and DC = 8 cm. If the area of ΔABC is 90 cm², find AB and AC.

    Tangent lengths: BD = BE = 10, CD = CF = 8, and let AE = AF = x. Area = ½ × r × perimeter = ½ × 4 × (2x + 36) = 90, so 2x + 36 = 45 and x = 4.5. AB = 4.5 + 10 = 14.5 cm and AC = 4.5 + 8 = 12.5 cm.

16. Important · 3 marks

    16. Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that ∠APB = 2∠OAB.

    PA = PB, so ∠PAB = ∠PBA = x, and ∠APB = 180° − 2x … (i) OA ⟂ PA, so ∠PAB + ∠OAB = 90°, giving x = 90° − ∠OAB … (ii) Substituting (ii) in (i): ∠APB = 180° − 2(90° − ∠OAB) = 2∠OAB.

17. Very Important · 4 marks

    17. A circular meditation garden has radius 9 m. From a point P, 18 m from the centre O, two tangents PA and PB are drawn to its boundary. (i) What is ∠OAP? (ii) Find the length of tangent PA. (iii) (A) Find ∠AOB. OR (B) If the tangents are inclined to each other at 60°, find the length of each tangent.

    (i) ∠OAP = 90° (the radius is perpendicular to the tangent). (ii) PA² = OP² − OA² = 324 − 81 = 243, so PA = 9√3 m. (iii) (A) cos ∠AOP = 9/18 = 1/2, so ∠AOP = 60° and ∠AOB = 120°. OR (B) ∠OPA = 30°, so tan 30° = 9/PA and PA = 9√3 m.