Class X Mathematics - Chapter 10: Circles

Correct Answer: (B) 100°

Explanation: In quadrilateral OAPB, ∠OAP = ∠OBP = 90° (tangent ⊥ radius). So ∠AOB = 360° - (90° + 90° + 80°) = 100°.

Source: CBSE Main 2024, Q12

Correct Answer: (A) 24 cm

Explanation: Using Pythagoras theorem: Length = √(25² - 7²) = √(625 - 49) = √576 = 24 cm.

Source: Compartment 2023, Q5

Correct Answer: (C) Tangent

Explanation: A line that touches a circle at exactly one point is called a tangent.

Source: CBSE Sample Paper 2025, Q8

Correct Answer: (B) 3√3 cm

Explanation: In ΔOAP, ∠OAP = 90°, ∠AOP = 30° (half of 60°). So tan30° = AP/OA ⇒ 1/√3 = AP/3 ⇒ AP = 3√3 cm.

Source: CBSE Main 2023, Q7

Correct Answer: (B) 2

Explanation: From any external point, exactly two tangents can be drawn to a circle.

Source: Compartment 2022, Q3

Correct Answer: (A) 12 cm

Explanation: In ΔOPQ, ∠OPQ = 90°, so PQ = √(OQ² - OP²) = √(169 - 25) = √144 = 12 cm.

Source: NCERT Exemplar, Ch 10, Q1

Correct Answer: (A) 60 cm²

Explanation: Area of PQOR = 2 × Area of ΔOPQ = 2 × (1/2 × PQ × OP) = PQ × OP. PQ = √(13² - 5²) = 12 cm. So area = 12 × 5 = 60 cm².

Source: CBSE Main 2022, Q9

Correct Answer: (B) 6 cm

Explanation: Let AB be chord of larger circle tangent to smaller circle at P. Then OP ⊥ AB and OP = 4 cm, OA = 5 cm. So AP = √(25 - 16) = 3 cm. Thus AB = 2 × 3 = 6 cm.

Source: NCERT Exemplar, Ch 10, Q5

Correct Answer: (C) Tangent

Explanation: A line that touches a circle at exactly one point is a tangent.

Source: CBSE Sample Paper 2024, Q10

Correct Answer: (B) 8 cm

Explanation: Length = √(10² - 6²) = √(100 - 36) = √64 = 8 cm.

Source: Compartment 2021, Q4

Correct Answer: (D) 55°

Explanation: In quadrilateral ABCD, ∠A + ∠B + ∠C + ∠D = 360°. Since ∠A = ∠B = 90° (tangents), then 90 + 90 + 125 + ∠COD = 360 ⇒ ∠COD = 55°.

Source: CBSE Main 2021, Q11

Correct Answer: (B) 50°

Explanation: In quadrilateral formed by two radii and two tangents, sum of angles = 360°. So angle between tangents = 360 - (90 + 90 + 130) = 50°.

Source: Compartment 2020, Q6

Correct Answer: (B) 2a

Explanation: In ΔOAP, sin30° = OA/OP ⇒ 1/2 = a/OP ⇒ OP = 2a.

Source: NCERT Exemplar, Ch 10, Q3

Correct Answer: (C) always less than OP

Explanation: In right triangle OAP, OP is hypotenuse and PA is perpendicular, so PA < OP.

Source: CBSE Main 2020, Q8

Correct Answer: (B) r

Explanation: In ΔOAP, ∠AOP = 45°, so tan45° = AP/OA ⇒ 1 = AP/r ⇒ AP = r.

Source: Compartment 2024, Q7

Correct Answer: (B) Tangent

Explanation: The road is tangent to the wheel at the point of contact.

Source: CBSE Sample Paper 2023, Q12

Correct Answer: (B) OP = 2AP

Explanation: In ΔOAP, ∠APO = 60°, so sin60° = OA/OP and cos60° = AP/OP. Thus 1/2 = AP/OP ⇒ OP = 2AP.

Source: CBSE Main 2025 (Expected Pattern), Q9

Correct Answer: (B) 2

Explanation: A circle can have exactly two parallel tangents (one at top and one at bottom).

Source: NCERT Textbook, Ch 10, Ex 10.1, Q3

Correct Answer: (A) one

Explanation: By definition, a tangent touches a circle at exactly one point.

Source: NCERT Textbook, Ch 10, Ex 10.1, Q1

Correct Answer: (C) secant

Explanation: A line that intersects a circle at two points is called a secant.

Source: NCERT Textbook, Ch 10, Ex 10.1, Q2

Model Answer:

Let AB be a diameter of a circle with center O. Let tangents PQ and RS be drawn at points A and B respectively.

∠OAP = 90° (tangent ⊥ radius) ...(1 mark)

∠OBR = 90° (tangent ⊥ radius) ...(1 mark)

Since ∠OAP = ∠OBR and they are alternate interior angles for lines PQ and RS with transversal AB, therefore PQ || RS.

Source: CBSE Main 2022, Q23

Model Answer:

Given AB = AC, so ΔABC is isosceles with ∠ABC = ∠ACB.

Tangents from external point to circle are equal: ...(1 mark)

BD = BE and CD = CE

Since AB = AC, then AB - BD = AC - CD ⇒ AD = AE

But AD = AE (tangents from A), so BE = EC ...(2 marks)

Source: Compartment 2021, Q28

Model Answer:

Let AB be the chord of larger circle touching smaller circle at P.

OP ⊥ AB (radius ⊥ tangent) ...(1 mark)

In ΔOPA, OA = 5 cm, OP = 3 cm

AP = √(OA² - OP²) = √(25 - 9) = √16 = 4 cm

So AB = 2 × AP = 8 cm ...(1 mark)

Source: CBSE Main 2023, Q24

Model Answer:

Let ABCD be a parallelogram circumscribing a circle.

AB + CD = AD + BC (sum of opposite sides of circumscribed quadrilateral) ...(1 mark)

But AB = CD and AD = BC (properties of parallelogram) ...(1 mark)

So 2AB = 2AD ⇒ AB = AD

Thus all sides equal, so ABCD is a rhombus ...(1 mark)

Source: NCERT Textbook, Ch 10, Ex 10.2, Q11

Model Answer:

For a quadrilateral circumscribing a circle: AB + CD = AD + BC ...(1 mark)

6 + 4 = AD + 7

10 = AD + 7

AD = 3 cm ...(1 mark)

Source: CBSE Main 2021, Q25

Model Answer:

Let PA and PB be tangents from external point P to circle with center O.

∠OAP = ∠OBP = 90° ...(1 mark)

In quadrilateral OAPB:

∠APB + ∠AOB + 90° + 90° = 360°

∠APB + ∠AOB = 180° ...(2 marks)

Source: NCERT Textbook, Ch 10, Ex 10.2, Q10

Model Answer:

In ΔOPQ, ∠OPQ = 90° (tangent ⊥ radius) ...(1 mark)

PQ = √(OQ² - OP²) = √(144 - 25) = √119 cm ...(1 mark)

Source: Compartment 2022, Q26

Model Answer:

In quadrilateral OPTQ:

∠OPT = ∠OQT = 90° (tangents) ...(1 mark)

∠PTQ = 360° - (90° + 90° + 110°) = 70° ...(1 mark)

Source: CBSE Main 2020, Q24

Model Answer:

Let PA be the tangent. In ΔOAP:

∠APO = 30° (half of 60°) ...(1 mark)

tan30° = OA/PA ⇒ 1/√3 = 3/PA ⇒ PA = 3√3 cm ...(1 mark)

Source: Compartment 2023, Q27

Model Answer:

Let XY be a tangent at point P to circle with center O.

Join OP. Then ∠OPX = 90° ...(1 mark)

Let PQ be perpendicular to XY at P.

Then ∠QPX = 90° ...(1 mark)

So ∠OPX = ∠QPX, which is possible only if O lies on PQ.

Thus perpendicular at P passes through center O ...(1 mark)

Source: NCERT Textbook, Ch 10, Ex 10.2, Q5

Model Answer:

Join OC. In ΔAOP and ΔAOC:

OP = OC (radii), OA = OA (common), ∠OPA = ∠OCA = 90°

So ΔAOP ≅ ΔAOC (RHS) ⇒ ∠1 = ∠2 ...(1 mark)

Similarly, ΔBOQ ≅ ΔBOC ⇒ ∠3 = ∠4 ...(1 mark)

Now, ∠1 + ∠2 + ∠3 + ∠4 = 180° (linear pair)

2(∠2 + ∠3) = 180° ⇒ ∠2 + ∠3 = 90° ⇒ ∠AOB = 90° ...(1 mark)

Source: NCERT Textbook, Ch 10, Ex 10.2, Q9

Model Answer:

Let AE = AF = x cm

Then AB = x + 8, AC = x + 6, BC = 14 ...(1 mark)

Area of ΔABC = √[s(s-a)(s-b)(s-c)] where s = (AB+BC+CA)/2 = (x+8+14+x+6)/2 = x+14

Also, area = r × s = 4(x+14) ...(1 mark)

Equating: 4(x+14) = √[(x+14)(6)(8)(x)]

Solving: x = 7, so AB = 15 cm, AC = 13 cm ...(1 mark)

Source: NCERT Textbook, Ch 10, Ex 10.2, Q12

Model Answer:

Let ABCD circumscribe a circle with center O.

Join OA, OB, OC, OD.

In ΔOAP and ΔOAS: AP = AS, OP = OS, OA = OA

So ΔOAP ≅ ΔOAS ⇒ ∠1 = ∠8 ...(1 mark)

Similarly, ∠2 = ∠3, ∠4 = ∠5, ∠6 = ∠7 ...(1 mark)

Sum of angles around O: ∠1+∠2+...+∠8 = 360°

2(∠1+∠2+∠5+∠6) = 360° ⇒ ∠AOB + ∠COD = 180°

Similarly, ∠BOC + ∠DOA = 180° ...(1 mark)

Source: NCERT Exemplar, Ch 10, Q14

Model Answer:

∠ABC = 90° (angle in semicircle) ...(1 mark)

∠BAC = 180° - (90° + 50°) = 40°

∠BAT = ∠ACB = 50° (alternate segment theorem) ...(1 mark)

Source: Compartment 2024, Q25

Model Answer:

Join OC and OD.

∠OBC = ∠OBD = 60° (tangents equally inclined) ...(1 mark)

In ΔOBC, cos60° = BC/BO ⇒ 1/2 = BC/BO ⇒ BO = 2BC ...(1 mark)

But BC = BD, so BC + BD = 2BC = BO ...(1 mark)

Source: NCERT Exemplar, Ch 10, Q10

Model Answer:

∠ACB = 90° (angle in semicircle) ...(1 mark)

∠ABC = 60° (angle sum)

∠BCD = ∠BAC = 30° (alternate segment theorem) ...(1 mark)

∠BDC = 180° - (60° + 30°) = 90°

In ΔBCD, ∠BCD = ∠BDC = 30°? Wait, recalculate:

Actually, ∠OCA = ∠OAC = 30° (OA=OC), so ∠OCD = 90°

∠BCD = ∠OCD - ∠OCB = 90° - 60° = 30°

∠BDC = 180° - 120° - 30° = 30°? Better approach:

∠COD = 2∠CAD = 2×30° = 60° (angle at center)

So ∠BDC = 30°, and ∠BCD = 30°, so BC = BD ...(1 mark)

Source: NCERT Exemplar, Ch 10, Q12

Model Answer:

PA = PB (tangents from external point) ...(1 mark)

∠PAB = (180° - 50°)/2 = 65°

∠OAB = 90° - 65° = 25° (since ∠OAP = 90°) ...(1 mark)

Source: CBSE Main 2025 (Expected Pattern), Q26

Model Answer:

Let PA and PB be tangents from external point P to circle with center O.

∠OAP = ∠OBP = 90° ...(1 mark)

In quadrilateral OAPB:

∠APB + ∠AOB + 90° + 90° = 360°

∠APB + ∠AOB = 180° ...(2 marks)

Source: NCERT Textbook, Ch 10, Ex 10.2, Q10 (Repeated for emphasis)

Model Answer:

Let the circle touch AB, BC, CD, DA at P, Q, R, S respectively.

AP = AS, BP = BQ, CQ = CR, DR = DS ...(1 mark)

DS = 5 cm ⇒ DR = 5 cm

AD = 23 cm ⇒ AS = 23 - 5 = 18 cm ⇒ AP = 18 cm

AB = 29 cm ⇒ BP = 29 - 18 = 11 cm ...(1 mark)

Since ∠B = 90°, OPBQ is a square ⇒ r = BP = 11 cm ...(1 mark)

Source: Compartment 2020, Q29

Model Answer:

TP = TQ (tangents from external point) ...(1 mark)

∠TPQ = ∠TQP

∠PTQ = 180° - 2∠TPQ

∠OPQ = 90° - ∠TPQ (since ∠OPT = 90°)

So 2∠OPQ = 180° - 2∠TPQ = ∠PTQ ...(1 mark)

Source: NCERT Textbook, Ch 10, Ex 10.2, Q8

Model Answer:

In ΔTPQ, TP = TQ (tangents from external point) ...(1 mark)

So ∠TPQ = ∠TQP

∠PTQ + ∠TPQ + ∠TQP = 180°

⇒ ∠PTQ = 180° - 2∠TPQ ...(1 mark)

Also, ∠OPT = 90° (tangent ⊥ radius)

So ∠OPQ = 90° - ∠TPQ ...(1 mark)

Now, 2∠OPQ = 2(90° - ∠TPQ) = 180° - 2∠TPQ = ∠PTQ

Therefore, ∠PTQ = 2∠OPQ ...(1 mark)

Source: CBSE Main 2023, Q31

Model Answer:

∠AOQ = 58° (given)

∠ABQ = (1/2)∠AOQ = 29° (angle at circumference is half of angle at center) ...(1 mark)

∠BAT = 90° (tangent ⊥ radius) ...(1 mark)

In ΔABT, ∠ABT + ∠BAT + ∠ATB = 180°

29° + 90° + ∠ATB = 180°

∠ATB = 61° ...(1 mark)

But ∠ATQ = ∠ATB = 61° (same angle) ...(1 mark)

Source: Compartment 2024, Q32

Model Answer:

Let AB be the tangent at R, so AB || PQ.

∠ARP = ∠RPQ (alternate interior angles) ...(1 mark)

But ∠ARP = ∠RQP (angle between tangent and chord equals angle in alternate segment) ...(1 mark)

So ∠RPQ = ∠RQP

Thus RP = RQ (sides opposite equal angles) ...(1 mark)

Therefore, R bisects arc PRQ (equal chords subtend equal arcs) ...(1 mark)

Source: NCERT Exemplar, Ch 10, Q15

Model Answer:

Produce AB and CD to meet at P.

For the first circle: PA = PC (tangents from P) ...(1 mark)

For the second circle: PB = PD (tangents from P) ...(1 mark)

Subtracting: PA - PB = PC - PD ...(1 mark)

So AB = CD ...(1 mark)

Source: NCERT Textbook, Ch 10, Ex 10.2, Q13

Model Answer:

∠AOQ = 58° (given)

∠ABQ = (1/2)∠AOQ = 29° (angle at circumference) ...(1 mark)

∠BAT = 90° (tangent ⊥ radius) ...(1 mark)

In ΔABT: ∠ABT + ∠BAT + ∠ATB = 180°

29° + 90° + ∠ATB = 180° ⇒ ∠ATB = 61° ...(1 mark)

∠ATQ = ∠ATB = 61° (same angle) ...(1 mark)

Source: Compartment 2024, Q32 (Repeated for emphasis)

Model Answer:

Let P be the midpoint of arc AB, so arc AP = arc PB.

Thus ∠ABP = ∠BAP (angles in equal arcs) ...(1 mark)

So ΔABP is isosceles with AP = BP.

Join OP. Then OP bisects ∠APB (property of isosceles triangle) ...(1 mark)

Also, OP ⊥ tangent at P (radius ⊥ tangent) ...(1 mark)

Since OP is perpendicular to both the tangent and the chord AB (as it bisects AB in isosceles triangle), the tangent is parallel to AB ...(1 mark)

Source: NCERT Exemplar, Ch 10, Q16

Model Answer:

Let AC = x, so AB = 2x.

In ΔOEC: OE = q, EC = x/2, OC = r

So q² + (x/2)² = r² ⇒ x² = 4(r² - q²) ...(1 mark)

In ΔOFB: OF = p, FB = x, OB = r

So p² + x² = r² ...(1 mark)

Substitute x²: p² + 4(r² - q²) = r² ...(1 mark)

p² + 4r² - 4q² = r² ⇒ 4q² = p² + 3r² ...(1 mark)

Source: NCERT Exemplar, Ch 10, Q18

Model Answer:

Let the circle touch AB, AC, BC at Q, R, P respectively.

AQ = AR, BQ = BP, CP = CR (tangents from external points) ...(1 mark)

Perimeter of ΔABC = AB + BC + CA

= (AQ - BQ) + (BP + PC) + (AR - CR) ...(1 mark)

= (AQ - BP) + (BP + CR) + (AQ - CR) (since BQ=BP, AR=AQ, CP=CR) ...(1 mark)

= AQ - BP + BP + CR + AQ - CR = 2AQ

Thus AQ = 1/2 (perimeter of ΔABC) ...(1 mark)

Source: NCERT Textbook, Ch 10, Ex 10.2, Q12 (Extended)

Model Answer:

Let PQ and RS be two parallel tangents to a circle with center O.

Let AB be the intercept of another tangent between them, touching at C.

Join OA, OB, OC.

In ΔOAP and ΔOAC: AP = AC, OP = OC, OA = OA

So ΔOAP ≅ ΔOAC ⇒ ∠1 = ∠2 ...(1 mark)

Similarly, ΔOBQ ≅ ΔOBC ⇒ ∠3 = ∠4 ...(1 mark)

Now, ∠1 + ∠2 + ∠3 + ∠4 = 180° (linear pair)

2(∠2 + ∠3) = 180° ⇒ ∠2 + ∠3 = 90° ...(1 mark)

In ΔAOB: ∠AOB = 180° - (∠2 + ∠3) = 90° ...(1 mark)

Source: NCERT Exemplar, Ch 10, Q13

Model Answer:

In quadrilateral OAPB:

∠OAP = ∠OBP = 90° (tangents) ...(1 mark)

∠AOB = 360° - (90° + 90° + 60°) = 120° ...(1 mark)

Now, if OA = AP:

In ΔOAP, ∠OAP = 90°, OA = AP

So ∠AOP = 45° (isosceles right triangle) ...(1 mark)

Similarly, ∠BOP = 45°

Thus ∠AOB = 90°, but this contradicts earlier result. Correction:

Actually, if OA = AP, then in ΔOAP:

tan(∠AOP) = AP/OA = 1 ⇒ ∠AOP = 45°

But ∠AOB = 2∠AOP = 90° only if... Wait, let's re-express:

Given OA = AP, and ∠OAP = 90°, then by Pythagoras: OP = √2 OA

But in ΔAOB: OA = OB (radii), ∠AOB = 120°

So AB² = OA² + OB² - 2·OA·OB·cos120° = 2r² - 2r²(-1/2) = 3r²

For equilateral, AB = OA = r, but √3 r ≠ r. So the condition must be different.

Correct approach: If ∠APB = 60°, then ∠AOB = 120° as above.

For ΔAOB to be equilateral, we need ∠AOB = 60°, which would require ∠APB = 120°.

Thus the question likely has a typo. Assuming intended: If ∠APB = 120°, then ∠AOB = 60°, and if OA = AP, then...

But to answer as per common interpretation:

∠AOB = 120° ...(1 mark)

For second part: If OA = AP, then in ΔOAP, ∠AOP = 45°, so ∠AOB = 90°, but this contradicts. So the condition for equilateral is different.

Actually, ΔAOB is equilateral only if ∠AOB = 60°, which requires ∠APB = 120°.

So with given ∠APB = 60°, ΔAOB cannot be equilateral. ...(1 mark)

Source: Adapted from Compartment 2022, Q33