CBSE Class 10 Mathematics Circles MCQs Set 13

Multiple Choice Questions (MCQs) for Class 10 Mathematics: Chapter 10 Circles

Explore reliable objective questions for Chapter 10 Circles tailored for Class 10 learners. Utilizing these Mathematics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Practice Chapter 10 Circles MCQs for Class 10 Mathematics

Access the complete set of multiple-choice questions for Chapter 10 Circles below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.

Multiple Choice Questions

Question 1: ABCD is a quadrilateral with ∠D = 90°. A circle with centre O touches AB, BC, CD and DA at P, Q, R and S. If BC = 38 cm, CD = 25 cm and BP = 27 cm, the radius of the circle is:
(a) 14 cm
(b) 11 cm
(c) 27 cm
(d) 25 cm
Show Answer & Explanation

Answer: (a) 14 cm

Explanation:
1. BQ = BP = 27, so CQ = 38 − 27 = 11 and CR = CQ = 11. Then DR = 25 − 11 = 14. ORDS is a square (right angle at D, OR = OS), so r = DR = 14 cm.

Teacher's Note:
1. Use equal tangent lengths from B, C and D.
2. ORDS is a square.

Question 2: Two tangents BC and BD are drawn from B to a circle with centre O, and ∠CBD = 120°. The ratio BO : BC is:
(a) 1 : 2
(b) √3 : 1
(c) 1 : 1
(d) 2 : 1
Show Answer & Explanation

Answer: (d) 2 : 1

Explanation:
1. BO bisects the angle, so ∠CBO = 60° and OC ⟂ BC. In the right triangle OCB, BC = BO cos 60° = BO/2, so BO = 2 BC.

Teacher's Note:
1. The line to the centre bisects the angle between the tangents.

Question 3: Circles are drawn with centres at the three vertices of a triangle with sides 4 cm, 6 cm and 8 cm so that each pair of circles touches externally. The radii of the circles are:
(a) 2 cm, 4 cm and 2 cm
(b) 3 cm, 4 cm and 2 cm
(c) 3 cm, 5 cm and 1 cm
(d) 4 cm, 4 cm and 2 cm
Show Answer & Explanation

Answer: (c) 3 cm, 5 cm and 1 cm

Explanation:
1. Let the radii be x, y, z with x + y = 4, y + z = 6 and x + z = 8. Adding: x + y + z = 9, so z = 5, x = 3, y = 1. (The radii 3 and 1 sum to 4, 1 and 5 to 6, 3 and 5 to 8.)

Teacher's Note:
1. For externally touching circles the distance between centres is the sum of the radii.
2. Solve three linear equations.

Question 4: The incircle of triangle ABC touches BC, CA and AB at P, Q and R. The sum AR + BP + CQ is equal to:
(a) the perimeter of triangle ABC
(b) half the perimeter of triangle ABC
(c) one third of the perimeter
(d) the semi-perimeter minus AB
Show Answer & Explanation

Answer: (b) half the perimeter of triangle ABC

Explanation:
1. AR = AQ, BP = BR, CP = CQ. So AR + BP + CQ = AQ + CP + BR, and the two groups together make the perimeter. Each group is half the perimeter.

Teacher's Note:
1. Equal tangents from each vertex.

Question 5: A circle touches side BC of triangle ABC and touches AB and AC produced at Q and R. The length AQ equals:
(a) the perimeter of triangle ABC
(b) half the perimeter of triangle ABC
(c) AB + AC − BC
(d) BC
Show Answer & Explanation

Answer: (b) half the perimeter of triangle ABC

Explanation:
1. AQ = AR, BQ = BP and CR = CP, where P is on BC. The perimeter is AB + BC + CA = AB + BP + PC + CA = AQ + AR = 2 AQ, so AQ is half the perimeter.

Teacher's Note:
1. Add the tangent lengths along the sides.

Question 6: AB is a line segment with midpoint M. On one side of AB, semicircles are drawn on AM, MB and AB as diameters. A circle is drawn that touches all three semicircles. Its radius is what fraction of AB?
(a) \( \frac16 \)
(b) \( \frac14 \)
(c) \( \frac13 \)
(d) \( \frac18 \)
Show Answer & Explanation

Answer: (a) \( \frac16 \)

Explanation:
1. Let AB = 2a. The small semicircles have radius a/2 and the large one radius a. The circle's centre is at height h above M, with h = a − r (internal contact with the large semicircle) and h² + a²/4 = (a/2 + r)² (external contact with a small one). Substituting: a² − 2ar = ar, so r = a/3 = AB/6.

Teacher's Note:
1. Place M at the origin and use the distances between centres.

Question 7: The locus of the midpoints of equal chords of a circle with centre O is:
(a) a straight line
(b) a circle with centre different from O
(c) a circle intersecting the given circle at the ends of the chords
(d) a circle with centre O
Show Answer & Explanation

Answer: (d) a circle with centre O

Explanation:
1. A chord of length l has its midpoint at distance √(r² − l²/4) from O, the same for all equal chords. So the midpoints lie on a circle concentric with the given one.

Teacher's Note:
1. Use the perpendicular from the centre to a chord.

Question 8: Two equal circles of radius 4 cm intersect so that each passes through the centre of the other. The length of the common chord is:
(a) \( 2\sqrt3 \) cm
(b) \( 4\sqrt2 \) cm
(c) \( 4\sqrt3 \) cm
(d) 8 cm
Show Answer & Explanation

Answer: (c) \( 4\sqrt3 \) cm

Explanation:
1. The centres are 4 cm apart. The chord is perpendicular to the line of centres at its midpoint, 2 cm from each centre, so half the chord is √(16 − 4) = 2√3. The chord is 4√3 cm.

Teacher's Note:
1. Draw the line of centres and the common chord.

Question 9: ABCD is a quadrilateral circumscribed about a circle of radius 10 cm, touching AB, BC, CD, DA at Q, R, S, P, with ∠DAB = 90°. If CS = 27 cm and CB = 38 cm, then AB is:
(a) 17 cm
(b) 28 cm
(c) 21 cm
(d) 19 cm
Show Answer & Explanation

Answer: (c) 21 cm

Explanation:
1. CR = CS = 27, so BR = 38 − 27 = 11 and BQ = 11. APOQ is a square, so AQ = 10. AB = AQ + QB = 21 cm.

Teacher's Note:
1. Equal tangent lengths.
2. Square formed at the right angle.

Question 10: AD and AE are tangents to a circle at D and E, and BC (with B on AD and C on AE) is a tangent at F. Then:
(a) AD = AB + BC + CA
(b) 2AD = AB + BC + CA
(c) 4AD = AB + BC + CA
(d) 3AD = AB + BC + CA
Show Answer & Explanation

Answer: (b) 2AD = AB + BC + CA

Explanation:
1. BD = BF and CE = CF. Perimeter = AB + BC + CA = AB + BD + CE + CA = AD + AE = 2 AD.

Teacher's Note:
1. Equal tangents.

Question 11: Assertion: If XY is a tangent to a circle with centre O touching at M, then OM ⟂ XY. Reason: The shortest distance from a point to a line is the perpendicular distance. Which is correct?
(a) Both are true and the Reason explains the Assertion
(b) Both are true but the Reason does not explain the Assertion
(c) Assertion is true but Reason is false
(d) Assertion is false but Reason is true
Show Answer & Explanation

Answer: (a) Both are true and the Reason explains the Assertion

Explanation:
1. Every point of the tangent other than M lies outside the circle, so OM is the shortest segment from O to the line, and the shortest segment to a line is the perpendicular. That is exactly the proof of the theorem.

Teacher's Note:
1. The proof of the tangent-radius theorem uses the shortest-distance property.

Question 12: A circle is inscribed in triangle ABC with AB = 12 cm, BC = 8 cm and CA = 10 cm, touching AB at D, BC at E and CA at F. The length BE is:
(a) 7 cm
(b) 3 cm
(c) 4 cm
(d) 5 cm
Show Answer & Explanation

Answer: (d) 5 cm

Explanation:
1. Semi-perimeter = 15. The tangent length from B is s − CA = 15 − 10 = 5 cm.

Teacher's Note:
1. Tangent length from a vertex = semi-perimeter minus the opposite side.

Question 13: A chord of a circle is equal to its radius. The angle it subtends at a point on the minor arc is:
(a) 60°
(b) 75°
(c) 120°
(d) 150°
Show Answer & Explanation

Answer: (d) 150°

Explanation:
1. The chord and the two radii form an equilateral triangle, so the central angle is 60°. The angle on the major arc is 30° and on the minor arc is 180° − 30° = 150°.

Teacher's Note:
1. Opposite angles of a cyclic quadrilateral add to 180°.

Question 14: Three circles of radius 20 have centres P, Q, R and overlap pairwise. Along the sides of triangle PQR the overlap of the circles on the three sides measures 5, 10 and 12. The perimeter of triangle PQR is:
(a) 120
(b) 66
(c) 93
(d) 87
Show Answer & Explanation

Answer: (c) 93

Explanation:
1. Each side equals the sum of the two radii minus the overlap: 40 − 5, 40 − 10 and 40 − 12. Perimeter = 120 − 27 = 93.

Teacher's Note:
1. Side = 40 minus overlap.

Question 15: AB and CD are parallel chords of a circle on the same side of the centre, AB = 8 cm, CD = 10 cm, and the distance between them is 2 cm. The radius is:
(a) \( \frac{4\sqrt{17}}{5} \) cm
(b) \( \frac{5\sqrt{17}}{4} \) cm
(c) \( \frac{3\sqrt{17}}{5} \) cm
(d) \( \sqrt{17} \) cm
Show Answer & Explanation

Answer: (b) \( \frac{5\sqrt{17}}{4} \) cm

Explanation:
1. Let the distances from the centre be d (to CD) and d + 2 (to AB). r² = 25 + d² = 16 + (d + 2)². So 25 + d² = 20 + 4d + d², d = 5/4, and r² = 25 + 25/16 = 425/16, r = 5√17/4. (Chords on opposite sides give a negative distance and are impossible.)

Teacher's Note:
1. Use the perpendicular from the centre to each chord.

Question 16: In the same set-up (DC = 12, DB = 9, BF = 3, DF = 6√2), the area of triangle BDF is:
(a) \( 9\sqrt2 \) cm²
(b) \( 3\sqrt2 \) cm²
(c) 161 cm²
(d) \( 18\sqrt2 \) cm²
Show Answer & Explanation

Answer: (a) \( 9\sqrt2 \) cm²

Explanation:
1. The triangle is right-angled at F, so area = ½ × DF × BF = ½ × 6√2 × 3 = 9√2 cm².

Teacher's Note:
1. Right angle at the point of contact.

Question 17: From a circular sheet of paper of radius 20 cm, four circles of radius 5 cm each are cut out. The ratio of the uncut portion to the cut portion is:
(a) 3 : 1
(b) 1 : 3
(c) 4 : 1
(d) 2 : 1
Show Answer & Explanation

Answer: (a) 3 : 1

Explanation:
1. Cut area = 4 × 25π = 100π. Total = 400π. Uncut = 300π. The ratio is 300 : 100 = 3 : 1.

Teacher's Note:
1. Compare the areas.

Question 18: A rectangle with sides 8 cm and 6 cm is inscribed in a circle. The difference between the area of the circle and the area of the rectangle is (π = 22/7):
(a) 78.57 cm²
(b) 28.57 cm²
(c) 32.57 cm²
(d) 30.57 cm²
Show Answer & Explanation

Answer: (d) 30.57 cm²

Explanation:
1. The diagonal 10 cm is a diameter, so r = 5 and the circle's area is 22/7 × 25 = 78.57. Subtract 48 to get 30.57 cm².

Teacher's Note:
1. The diagonal of the rectangle is the diameter.

Question 19: ABCD is a square of side 14 cm. Circles of radius 7 cm are drawn with centres at A, B, C and D, so that each touches two of the others externally. The area of the part of the square not covered by any circle is (π = 22/7):
(a) 196 cm²
(b) 154 cm²
(c) 42 cm²
(d) 56 cm²
Show Answer & Explanation

Answer: (c) 42 cm²

Explanation:
1. Each circle contributes a quarter inside the square. Together they cover one full circle: 154 cm². Remaining = 196 − 154 = 42 cm².

Teacher's Note:
1. Four quarter circles.

Question 20: Two circles touch externally. The sum of their areas is \( 130\pi \) cm² and the distance between their centres is 14 cm. The radii are:
(a) 10 cm and 4 cm
(b) 11 cm and 3 cm
(c) 9 cm and 5 cm
(d) 12 cm and 2 cm
Show Answer & Explanation

Answer: (b) 11 cm and 3 cm

Explanation:
1. r1 + r2 = 14 and r1² + r2² = 130. Then 2 r1 r2 = 196 − 130 = 66, r1 r2 = 33, so the radii are roots of t² − 14t + 33 = 0, which gives 11 and 3.

Teacher's Note:
1. Use (r1 + r2)² = r1² + r2² + 2 r1 r2.

Question 21: PQRS is a diameter of a circle of radius 6 cm with PQ = QR = RS. Semicircles are drawn on PQ (below PS) and on QS (above PS). The shaded region is the upper half of the big circle outside the semicircle on QS, together with the semicircle on PQ. Its area is (π = 22/7):
(a) 25.14 cm²
(b) 37.71 cm² (that is, 12π)
(c) 18.86 cm²
(d) 56.57 cm²
Show Answer & Explanation

Answer: (b) 37.71 cm² (that is, 12π)

Explanation:
1. Upper half of the big circle = ½π(36) = 18π. Semicircle on QS (diameter 8) = ½π(16) = 8π. Semicircle on PQ (diameter 4) = ½π(4) = 2π. Area = 18π − 8π + 2π = 12π ≈ 37.71 cm².

Teacher's Note:
1. Add and subtract semicircle areas.

Question 22: For the same shaded region (PQRS a diameter of the circle of radius 6 cm, PQ = QR = RS), the perimeter is:
(a) 12π cm
(b) 10π cm
(c) 18π cm
(d) 6π cm
Show Answer & Explanation

Answer: (a) 12π cm

Explanation:
1. The boundary is the arc of the big semicircle (6π), the arc of the semicircle on QS (4π) and the arc of the semicircle on PQ (2π). Total = 12π cm.

Teacher's Note:
1. Add the three arcs.

Question 23: AOB is a quadrant of a circle of radius 3.5 cm with centre O. D is a point on OB with OD = 2 cm. The area of the region bounded by DA, the arc AB and BD is (π = 22/7):
(a) 9.625 cm²
(b) 3.5 cm²
(c) 5.25 cm²
(d) 6.125 cm²
Show Answer & Explanation

Answer: (d) 6.125 cm²

Explanation:
1. Quadrant area = ¼ × 22/7 × 3.5² = 9.625. Triangle OAD = ½ × 3.5 × 2 = 3.5. Region = 9.625 − 3.5 = 6.125 cm².

Teacher's Note:
1. Subtract the right triangle from the quadrant.

Question 24: A semicircle is drawn inside a square of side 10 cm on each of its four sides, using the side as diameter. The petal-shaped regions where the semicircles overlap (four petals meeting at the centre) have total area (π = 3.14):
(a) 100 cm²
(b) 78.5 cm²
(c) 57 cm²
(d) 157 cm²
Show Answer & Explanation

Answer: (c) 57 cm²

Explanation:
1. The four semicircles have total area 4 × ½ × π × 25 = 157. They cover the square once and the petals twice, so petals = 157 − 100 = 57 cm².

Teacher's Note:
1. Total of semicircle areas minus the square.

Question 25: AB and CD are perpendicular diameters of a circle of radius 7 cm with centre O, and OD is the diameter of a smaller circle. The shaded region is the small circle together with the two segments of the big circle cut off by chords BC and CA. Its area is (π = 22/7):
(a) 154 cm²
(b) 38.5 cm²
(c) 66.5 cm²
(d) 52.5 cm²
Show Answer & Explanation

Answer: (c) 66.5 cm²

Explanation:
1. Small circle: radius 3.5, area 38.5. Each segment = quadrant 38.5 − triangle 24.5 = 14, two of them 28. Total = 66.5 cm².

Teacher's Note:
1. Segment = quadrant minus triangle.

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