CBSE Class 10 Mathematics Circles MCQs Set 14

Download CBSE 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.

Chapter-wise Objective Questions: Chapter 10 Circles

View or download the dedicated Chapter 10 Circles MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.

Multiple Choice Questions

Question 1: From an external point P, tangents PA and PB are drawn to a circle with centre O. A tangent CD at a point E of the circle meets PA at C and PB at D. If PA = 14 cm, the perimeter of triangle PCD is:
(a) \( 28\text{ cm} \)
(b) \( 14\text{ cm} \)
(c) \( 21\text{ cm} \)
(d) \( 42\text{ cm} \)
Show Answer & Explanation

Answer: (a) \( 28\text{ cm} \)

Explanation:
1. Tangent lengths from C are CA = CE and from D are DB = DE. Perimeter = PC + CE + ED + DP = PC + CA + DB + DP = PA + PB = 14 + 14 = 28 cm.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 2: A circle is inscribed in triangle ABC with AB = 10 cm, BC = 12 cm and CA = 8 cm, touching the sides at D (on AB), E (on BC) and F (on CA). The lengths AD, BE, CF are respectively:
(a) \( 5, 5, 7\text{ cm} \)
(b) \( 4, 6, 6\text{ cm} \)
(c) \( 3, 5, 7\text{ cm} \)
(d) \( 3, 7, 5\text{ cm} \)
Show Answer & Explanation

Answer: (d) \( 3, 7, 5\text{ cm} \)

Explanation:
1. Let AD = AF = x, BD = BE = y, CE = CF = z. Then x + y = 10, y + z = 12, z + x = 8. Adding, x + y + z = 15, so x = 3, y = 7, z = 5.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 3: ABCD is a quadrilateral with angle D = 90 degrees. A circle with centre O and radius r touches the sides AB, BC, CD, DA at P, Q, R, S respectively. If BC = 38 cm, CD = 25 cm and BP = 27 cm, the value of r is:
(a) \( 11\text{ cm} \)
(b) \( 12\text{ cm} \)
(c) \( 14\text{ cm} \)
(d) \( 16\text{ cm} \)
Show Answer & Explanation

Answer: (c) \( 14\text{ cm} \)

Explanation:
1. BQ = BP = 27, so CQ = 38 - 27 = 11 and CR = CQ = 11. Then DR = 25 - 11 = 14. Since OR is perpendicular to CD, OS is perpendicular to DA and angle D = 90 degrees, ORDS is a square, so r = OR = DR = 14 cm.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 4: A point P is 7 cm from the centre of a circle of diameter 8 cm. The number of tangents that can be drawn from P to the circle is:
(a) 0
(b) 2
(c) 1
(d) Infinitely many
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. The radius is 4 cm and OP = 7 cm > 4 cm, so P lies outside the circle. Exactly two tangents can be drawn from an external point.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 5: The distance between two parallel tangents to a circle of radius 4.5 cm is:
(a) \( 4.5\text{ cm} \)
(b) \( 9\text{ cm} \)
(c) \( 13.5\text{ cm} \)
(d) \( 18\text{ cm} \)
Show Answer & Explanation

Answer: (b) \( 9\text{ cm} \)

Explanation:
1. Parallel tangents touch the circle at the two ends of a diameter, so the distance between them is the diameter = 2 x 4.5 = 9 cm.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 6: A square circumscribes a circle of radius 5 cm. The length of a diagonal of the square is:
(a) \( 10\sqrt2\text{ cm} \)
(b) \( 10\text{ cm} \)
(c) \( 5\sqrt2\text{ cm} \)
(d) \( 20\text{ cm} \)
Show Answer & Explanation

Answer: (a) \( 10\sqrt2\text{ cm} \)

Explanation:
1. The side of the square equals the diameter, 10 cm. The diagonal is 10√2 cm.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 7: PA and PB are tangents from P to a circle with centre C. If angle APB = 50 degrees, then angle ACB is:
(a) \( 50^\circ \)
(b) \( 90^\circ \)
(c) \( 100^\circ \)
(d) \( 130^\circ \)
Show Answer & Explanation

Answer: (d) \( 130^\circ \)

Explanation:
1. In quadrilateral ACBP, angles at A and B are 90 degrees each. So angle ACB = 360 - 90 - 90 - 50 = 130 degrees.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 8: PQ and PR are tangents from P to a circle with centre O. If angle QOP = 55 degrees, then angle QPR is:
(a) \( 35^\circ \)
(b) \( 55^\circ \)
(c) \( 70^\circ \)
(d) \( 110^\circ \)
Show Answer & Explanation

Answer: (c) \( 70^\circ \)

Explanation:
1. OP bisects angle QOR, so angle QOR = 2 x 55 = 110 degrees. In quadrilateral OQPR, angle QPR = 360 - 90 - 90 - 110 = 70 degrees.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 9: PA and PB are tangents from an external point P to a circle with centre C. If angle APB = 80 degrees, then angle ACP is:
(a) \( 40^\circ \)
(b) \( 80^\circ \)
(c) \( 50^\circ \)
(d) \( 100^\circ \)
Show Answer & Explanation

Answer: (c) \( 50^\circ \)

Explanation:
1. Angle ACB = 180 - 80 = 100 degrees, and CP bisects it, so angle ACP = 50 degrees.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 10: PA and PB are tangents from P to a circle with centre C and radius 4 cm, with PA perpendicular to PB. The length of each tangent is:
(a) \( 2\sqrt2\text{ cm} \)
(b) \( 4\text{ cm} \)
(c) \( 8\text{ cm} \)
(d) \( 4\sqrt2\text{ cm} \)
Show Answer & Explanation

Answer: (b) \( 4\text{ cm} \)

Explanation:
1. CAPB has three right angles (at A, B, P), so it is a rectangle with CA = CB = 4, hence a square. PA = CB = 4 cm.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 11: The length of the tangent drawn to a circle of radius 7 cm from a point 25 cm away from the centre is:
(a) \( 24\text{ cm} \)
(b) \( 18\text{ cm} \)
(c) \( 32\text{ cm} \)
(d) \( \sqrt{674}\text{ cm} \)
Show Answer & Explanation

Answer: (a) \( 24\text{ cm} \)

Explanation:
1. Tangent length = √(25² - 7²) = √(625 - 49) = √576 = 24 cm.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 12: A point P is 26 cm from the centre of a circle and the tangent from P to the circle has length 24 cm. The radius of the circle is:
(a) \( 8\text{ cm} \)
(b) \( 12\text{ cm} \)
(c) \( \sqrt{1252}\text{ cm} \)
(d) \( 10\text{ cm} \)
Show Answer & Explanation

Answer: (d) \( 10\text{ cm} \)

Explanation:
1. r² = OP² - PT² = 676 - 576 = 100, so r = 10 cm.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 13: Two concentric circles have radii 6 cm (outer) and 4 cm (inner). PA is a tangent to the outer circle at A and PB is a tangent to the inner circle at B, from the same point P with PA = 10 cm. PB, to one decimal place, is:
(a) \( 8.0\text{ cm} \)
(b) \( 9.2\text{ cm} \)
(c) \( 11.7\text{ cm} \)
(d) \( 10.9\text{ cm} \)
Show Answer & Explanation

Answer: (d) \( 10.9\text{ cm} \)

Explanation:
1. OP² = PA² + OA² = 100 + 36 = 136. PB² = OP² - OB² = 136 - 16 = 120, so PB = 10.95, about 10.9 cm.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 14: A quadrilateral ABCD is circumscribed about a circle. If BC = 7 cm, CD = 3 cm and AD = 5 cm, then AB is:
(a) \( 5\text{ cm} \)
(b) \( 8\text{ cm} \)
(c) \( 9\text{ cm} \)
(d) \( 15\text{ cm} \)
Show Answer & Explanation

Answer: (c) \( 9\text{ cm} \)

Explanation:
1. For a circumscribed quadrilateral AB + CD = BC + AD. So AB + 3 = 7 + 5, giving AB = 9 cm.

Teacher's Note:
1. In a quadrilateral circumscribed about a circle, the sums of opposite sides are equal.

Question 15: A quadrilateral ABCD is circumscribed about a circle. The tangent lengths from the vertices A, B, C, D are 6 cm, 5 cm, 3 cm and 4 cm respectively. The perimeter of ABCD is:
(a) \( 18\text{ cm} \)
(b) \( 36\text{ cm} \)
(c) \( 30\text{ cm} \)
(d) \( 32\text{ cm} \)
Show Answer & Explanation

Answer: (b) \( 36\text{ cm} \)

Explanation:
1. Each side is the sum of the tangent lengths from its two ends. The perimeter counts each tangent length twice: 2(6 + 5 + 3 + 4) = 36 cm.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 16: From an external point P, tangents PA and PB are drawn to a circle. A tangent at a point D of the circle meets PA at E and PB at C. If AP = 15 cm, the perimeter of triangle PEC is:
(a) \( 30\text{ cm} \)
(b) \( 15\text{ cm} \)
(c) \( 45\text{ cm} \)
(d) \( 22.5\text{ cm} \)
Show Answer & Explanation

Answer: (a) \( 30\text{ cm} \)

Explanation:
1. As in the standard result, the perimeter of triangle PEC = PE + EC + CP = PE + EA + CB + CP = PA + PB = 15 + 15 = 30 cm.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 17: ABC is a triangle right-angled at A with AB = 5 cm and BC = 13 cm. The radius of its incircle is:
(a) \( 2\text{ cm} \)
(b) \( 1.5\text{ cm} \)
(c) \( 2.5\text{ cm} \)
(d) \( 3\text{ cm} \)
Show Answer & Explanation

Answer: (a) \( 2\text{ cm} \)

Explanation:
1. AC = √(169 - 25) = 12. For a right triangle, r = (AB + AC - BC)/2 = (5 + 12 - 13)/2 = 2 cm.

Teacher's Note:
1. The inradius of a right-angled triangle is (sum of the legs - hypotenuse) / 2.

Question 18: A circle touches all four sides of quadrilateral ABCD with AB = 6 cm, BC = 7 cm and CD = 4 cm. The length of AD is:
(a) \( 5\text{ cm} \)
(b) \( 9\text{ cm} \)
(c) \( 17\text{ cm} \)
(d) \( 3\text{ cm} \)
Show Answer & Explanation

Answer: (d) \( 3\text{ cm} \)

Explanation:
1. AB + CD = BC + AD gives 6 + 4 = 7 + AD, so AD = 3 cm.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 19: PA and PB are tangents from P to a circle with PA = 9 cm and angle APB = 60 degrees. The length of chord AB is:
(a) \( 4.5\text{ cm} \)
(b) \( 9\sqrt3\text{ cm} \)
(c) \( 9\text{ cm} \)
(d) \( 18\text{ cm} \)
Show Answer & Explanation

Answer: (c) \( 9\text{ cm} \)

Explanation:
1. PA = PB, so angles PAB and PBA are equal; each is (180 - 60)/2 = 60 degrees. Triangle PAB is equilateral, so AB = 9 cm.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 20: Two tangents PA and PB are drawn to a circle with centre O and radius r from a point P with OP = 2r. Triangle APB is:
(a) isosceles but not equilateral
(b) equilateral
(c) right-angled at P
(d) scalene
Show Answer & Explanation

Answer: (b) equilateral

Explanation:
1. In right triangle OAP, sin(angle APO) = OA/OP = 1/2, so angle APO = 30 degrees and angle APB = 60 degrees. With PA = PB, the triangle is equilateral.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 21: PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q meet at T. The length PT is:
(a) \( 5\text{ cm} \)
(b) \( 6\tfrac23\text{ cm} \)
(c) \( 6\text{ cm} \)
(d) \( 8\text{ cm} \)
Show Answer & Explanation

Answer: (b) \( 6\tfrac23\text{ cm} \)

Explanation:
1. Let M be the midpoint of PQ. PM = 4, OM = √(25 - 16) = 3. In right triangle OPT with PM perpendicular to OT, OP² = OM x OT, so 25 = 3 x OT and OT = 25/3. PT² = OT² - OP² = 625/9 - 25 = 400/9, PT = 20/3 = 6 2/3 cm.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 22: ABC is a triangle right-angled at A with AB = 6 cm and AC = 8 cm. The radius of the circle inscribed in it is:
(a) \( 2\text{ cm} \)
(b) \( 3\text{ cm} \)
(c) \( 2.4\text{ cm} \)
(d) \( 4\text{ cm} \)
Show Answer & Explanation

Answer: (a) \( 2\text{ cm} \)

Explanation:
1. BC = 10. r = (6 + 8 - 10)/2 = 2 cm. (Check: area = 24 = r x s = r x 12.)

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 23: A chord of a circle has length 16 cm and is at a distance of 15 cm from the centre. The radius of the circle is:
(a) \( 15\text{ cm} \)
(b) \( 16\text{ cm} \)
(c) \( \sqrt{31}\text{ cm} \)
(d) \( 17\text{ cm} \)
Show Answer & Explanation

Answer: (d) \( 17\text{ cm} \)

Explanation:
1. Half the chord is 8. r² = 15² + 8² = 289, so r = 17 cm.

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

Question 24: The radius of a circle is 6 cm. The perpendicular distance from the centre to a chord of length 8 cm is:
(a) \( 2\text{ cm} \)
(b) \( 4\text{ cm} \)
(c) \( 2\sqrt5\text{ cm} \)
(d) \( \sqrt{52}\text{ cm} \)
Show Answer & Explanation

Answer: (c) \( 2\sqrt5\text{ cm} \)

Explanation:
1. Half the chord is 4. Distance = √(36 - 16) = √20 = 2√5 cm.

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

Question 25: An equilateral triangle ABC is inscribed in a circle with centre O. Then angle BOC is:
(a) \( 60^\circ \)
(b) \( 90^\circ \)
(c) \( 120^\circ \)
(d) \( 150^\circ \)
Show Answer & Explanation

Answer: (c) \( 120^\circ \)

Explanation:
1. Angle BAC = 60 degrees. The angle at the centre is twice the angle at the circumference, so angle BOC = 120 degrees.

Teacher's Note:
1. The angle subtended by an arc at the centre is double the angle it subtends at any point on the remaining circle.

Question 26: O is the centre of a circle through A, B, C where O lies inside triangle ABC. If angle OBC = 25 degrees, then angle BAC is:
(a) \( 25^\circ \)
(b) \( 65^\circ \)
(c) \( 50^\circ \)
(d) \( 130^\circ \)
Show Answer & Explanation

Answer: (b) \( 65^\circ \)

Explanation:
1. OB = OC, so angle OCB = 25 degrees and angle BOC = 180 - 50 = 130 degrees. Angle BAC = 130/2 = 65 degrees.

Teacher's Note:
1. The angle at the circumference is half the angle at the centre on the same arc.

Question 27: In a circle with centre O, AB and CD are two perpendicular diameters. The length of chord AC is:
(a) \( \dfrac{1}{\sqrt2} \) AB
(b) AB
(c) \( \dfrac{1}{2} \) AB
(d) \( \sqrt2 \) AB
Show Answer & Explanation

Answer: (a) \( \dfrac{1}{\sqrt2} \) AB

Explanation:
1. OA = OC = r and angle AOC = 90 degrees, so AC = r√2. AB = 2r, so AC = AB/√2.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 28: In a circle with centre O, two unequal chords AB and CD intersect at P inside the circle. Triangles APC and DPB are:
(a) congruent
(b) equal in area but not similar
(c) neither similar nor congruent
(d) similar
Show Answer & Explanation

Answer: (d) similar

Explanation:
1. Angle APC = angle DPB (vertically opposite) and angle CAB = angle CDB (angles in the same segment). By AA, the triangles are similar; they are not congruent because the chords are unequal.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

Question 29: Two circles touch externally at C. A direct common tangent touches them at A and B, and the common tangent at C meets AB at D. If DC = 4 cm, then AB is:
(a) \( 4\text{ cm} \)
(b) \( 6\text{ cm} \)
(c) \( 16\text{ cm} \)
(d) \( 8\text{ cm} \)
Show Answer & Explanation

Answer: (d) \( 8\text{ cm} \)

Explanation:
1. DA = DC and DB = DC (tangent lengths from D). So AB = DA + DB = 8 cm.

Teacher's Note:
1. Tangents drawn from an external point to a circle are equal in length and each is perpendicular to the radius at its point of contact.

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Practice MCQs for Class 10 Mathematics Chapter 10 Circles

Class 10 Mathematics Chapter 10 Circles Objective Test Questions

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FAQs

Where can I access latest CBSE Class 10 Mathematics Circles MCQs Set 14?

You can get most exhaustive CBSE Class 10 Mathematics Circles MCQs Set 14 for free on StudiesToday.com. These MCQs for Class 10 Mathematics are updated for the 2026-27 academic session as per CBSE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Mathematics Class 10 material?

Yes, our CBSE Class 10 Mathematics Circles MCQs Set 14 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the CBSE paper is now competency-based.

How do practicing Mathematics MCQs help in scoring full marks in Class 10 exams?

By solving our CBSE Class 10 Mathematics Circles MCQs Set 14, Class 10 students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Mathematics.

Do you provide answers and explanations for CBSE Class 10 Mathematics Circles MCQs Set 14?

Yes, Mathematics MCQs for Class 10 have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused CBSE exams.

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