IMO Maths Olympiad Class 10 Chapter 10 Circles MCQs with Answers

Answer: (a) x° + 2y° = 90°

Explanation:
1. By the tangent-chord (alternate segment) theorem, ∠QSR = ∠QRP = y°.
2. SQ is a diameter, so ∠SRQ = 90° (angle in a semicircle).
3. In △PRS: ∠SPR + ∠PSR + ∠PRS = 180°, i.e. x° + y° + (y° + 90°) = 180°, so x° + 2y° = 90°.

Teacher's Note:
1. The angle between a tangent and a chord equals the angle in the alternate segment.
2. ∠PRS is made of ∠QRP and the right angle ∠QRS.

Answer: (d) 3 cm

Explanation:
1. Tangents from a point to a circle are equal. So with R, Q, P, S as the points of contact, AR = AS, BR = BQ, CQ = CP and DP = DS.
2. Adding, AB + CD = BC + AD.
3. 6 + 4 = 7 + AD, so AD = 3 cm.

Teacher's Note:
1. In a quadrilateral that has an inscribed circle, the sums of opposite sides are equal.
2. Add the equal tangent pairs vertex by vertex.

Answer: (b) 10 cm

Explanation:
1. Tangents from the same point are equal: BP = BQ, CP = CR and AQ = AR = 5 cm.
2. Perimeter = AB + BC + CA = AB + (BP + PC) + CA = (AB + BQ) + (CR + CA) = AQ + AR.
3. So the perimeter = 5 + 5 = 10 cm.

Teacher's Note:
1. Replace BP by BQ and CP by CR; the perimeter then collapses into AQ + AR.
2. For an excircle, the perimeter of the triangle is twice the tangent from the opposite vertex.

Answer: (a) 25 cm

Explanation:
1. Tangents from a point are equal: AD = AF = x, BD = BE = y and CE = CF = z.
2. Then x + y = 10, y + z = 12 and x + z = 28. Adding, 2(x + y + z) = 50.
3. AD + BE + CF = x + y + z = 25 cm.

Teacher's Note:
1. The required sum is simply the semi-perimeter of the triangle.
2. The measures printed in the book do not satisfy the triangle inequality (10 + 12 < 28), but the working above uses only the tangent property, which is how the key arrives at 25.

Answer: (d) 2(∠XZY + ∠YXZ)

Explanation:
1. O and Y lie on opposite sides of the chord XZ, so Y is on the minor arc XZ. Then ∠XOZ = 360° − 2∠XYZ.
2. In △XYZ: ∠XYZ = 180° − (∠XZY + ∠YXZ).
3. So ∠XOZ = 360° − 2[180° − (∠XZY + ∠YXZ)] = 2(∠XZY + ∠YXZ).

Teacher's Note:
1. Decide first which arc the inscribed angle stands on before using 'central angle = 2 × inscribed angle'.
2. Here the centre and the point Y are on opposite sides of XZ.

Answer: (a) 60°, 150°

Explanation:
1. By the alternate segment theorem, the angle in the alternate segment ∠APB = ∠BAT = 30°.
2. The angle at the centre is double: ∠AOB = 2 × 30° = 60°.
3. APBQ is a cyclic quadrilateral, so ∠AQB = 180° − 30° = 150°.

Teacher's Note:
1. The tangent-chord angle equals the angle in the alternate segment.
2. Opposite angles of a cyclic quadrilateral add up to 180°.

Answer: (d) 50°

Explanation:
1. Radii OA and OB are perpendicular to the tangents, so ∠PAO = ∠PBO = 90°.
2. In quadrilateral APBO: ∠AOB = 360° − 90° − 90° − 80° = 100°.
3. The angle at the circumference is half the angle at the centre: ∠AQB = 100° ÷ 2 = 50°.

Teacher's Note:
1. The angle between the two tangents and the angle at the centre are supplementary.
2. Q is on the major arc, so the inscribed angle is half of 100°.

Answer: (c) 2 : 1

Explanation:
1. BA is a diameter, so ∠BPA = 90° and ∠PAB = 60°. Since OP = OA, △OPA is equilateral and AP = r.
2. OP ⊥ PT, so ∠APT = 90° − 60° = 30° and ∠ATP = 60° − 30° = 30°; hence AT = AP = r.
3. BA : AT = 2r : r = 2 : 1.

Teacher's Note:
1. An angle in a semicircle is a right angle.
2. Showing that triangle APT is isosceles gives AT = r.

Answer: (b) 150°

Explanation:
1. In the circle with centre B, the angle at the centre on the arc QY that does not contain X is 210°, so ∠YXQ = 210° ÷ 2 = 105°.
2. PXQ is a straight line, so ∠PXY = 180° − 105° = 75°.
3. In the circle with centre A, the angle at the centre is double: obtuse ∠PAY = 2 × 75° = 150°.

Teacher's Note:
1. Use the angle at the centre = 2 × the angle at the circumference twice, once in each circle.
2. A straight line gives a linear pair at X.

Answer: (b) (a) → (r), (b) → (q), (c) → (p)

Explanation:
1. (a) OC ⊥ AB bisects it, so CA = 3.5 and CP = 3.5 + 9 = 12.5. OC² = 13² − 12.5² = 12.75, and OA² = 12.75 + 3.5² = 25, so the radius is 5 cm → (r).
2. (b) OT ⊥ PT, so (x + 4)² = x² + 8². Then 8x = 48 and x = 6 → (q).
3. (c) PQ = PR = 10, and the tangent segments on the third side add up to PS + ST + TP = PQ + PR = 20 cm → (p).

Teacher's Note:
1. Tangents from one point are equal; the perimeter of the cut-off triangle is twice the tangent length from P.
2. Remember that BA is 7 cm in total, so each half is 3.5 cm.