Circles: 3 marks Questions (CBSE Class 10)

52 different 3 marks questions on Circles from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (100)2 marks (40)3 marks (52)4 marks (29)5 marks (10)

Q293 marksShort AnswerCircles CBSE 2026 · Standard 30/1/1

In the given figure, $△ A B C$ is a right triangle in which $∠ B = 9 0^{\circ}$ , $A B = 4$ cm and $B C = 3$ cm. Find the radius of the circle inscribed in the triangle ABC.

Diagram for CBSE 2026 Class 10 Maths question 29

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Answer: 1 cm

$A C = 4^{2} + 3^{2} = 5$ cm.
Let the incircle (centre O, radius $r$ ) touch BC, CA, AB at P, Q, R. OPBR is a square, so $B P = B R = r$ .
Tangent lengths: $C P = C Q = 3 - r$ and $A R = A Q = 4 - r$ .
$A C = A Q + QC$ : $5 = (4 - r) + (3 - r)$ , so $2 r = 2$ and $r = 1$ cm.

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q29 (OR) (OR)3 marksShort AnswerCircles CBSE 2026 · Standard 30/1/1

In the given figure, if a circle touches the side QR of $△ P QR$ at S and extended sides PQ and PR at M and N respectively, then prove that : $P M = \frac{1}{2} (P Q + QR + P R)$

Diagram for CBSE 2026 Class 10 Maths question 29 (OR)

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Answer: Proved.

Tangents from an external point are equal: $P M = P N$ , $QM = QS$ , $R N = R S$ .
Perimeter $P Q + QR + P R = P Q + (QS + S R) + P R = (P Q + QM) + (P R + R N) = P M + P N$ .
Since $P M = P N$ , perimeter $= 2 P M$ .
So $P M = \frac{1}{2} (P Q + QR + P R)$ .

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q313 marksShort AnswerCircles CBSE 2026 · Standard 30/2/2

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

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Answer: Proved.

Let XY be the tangent at point P of a circle with centre O. To prove: OP $⊥$ XY.
Take any point Q on XY other than P. Q lies outside the circle (a tangent meets the circle at only one point).
So OQ > radius = OP.
Thus OP is the shortest of all segments from O to points of XY.
The shortest segment from a point to a line is the perpendicular, so OP $⊥$ XY.

Q31 (OR) (OR)3 marksShort AnswerCircles CBSE 2026 · Standard 30/2/2

If a regular hexagon ABCDEF circumscribes a circle, then prove that AB + CD + EF = BC + DE + FA.

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Answer: Proved.

Let the sides AB, BC, CD, DE, EF, FA touch the circle at P, Q, R, S, T, U respectively.
Tangents from an external point are equal: AP = AU, BP = BQ, CR = CQ, DR = DS, ET = ES, FT = FU.
AB + CD + EF = (AP + PB) + (CR + RD) + (ET + TF)
= (AU + BQ) + (CQ + DS) + (ES + FU)
= (BQ + QC) + (DS + SE) + (FU + UA) = BC + DE + FA
Hence proved.

Q313 marksShort AnswerCircles CBSE 2026 · Standard 30/3/1

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

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Answer: Proved.

Let $∠ A P B = θ$
PA = PB (tangents from an external point), so $∠ P A B = ∠ P B A = \frac{18 0 ^{\circ} - θ}{2} = 9 0^{\circ} - \frac{θ}{2}$
$O A ⊥ P A$ , so $∠ O A P = 9 0^{\circ}$
$∠ O A B = 9 0^{\circ} - ∠ P A B = 9 0^{\circ} - (9 0^{\circ} - \frac{θ}{2}) = \frac{θ}{2}$
Hence $∠ A P B = 2∠ O A B$

Also asked in: 2026 Standard 30/3/2, 2026 Standard 30/3/3

Q31 (OR) (OR)3 marksShort AnswerCircles CBSE 2026 · Standard 30/3/1

In the given figure, PA is the tangent to the circle with centre O such that OA = 10 cm, AB = 8 cm and AB $⊥$ OP. Find the length of PB.

Diagram for CBSE 2026 Class 10 Maths question 31 (OR)

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Answer:

\frac{32}{3}

cm (about 10.67 cm)

In right $△ O B A$ : $O B = O A^{2} - A B^{2} = 100 - 64 = 6$ cm
$O A ⊥ P A$ , so $△ O A P$ is right-angled at A, and AB is the altitude to OP
$△ A B P \sim △ O B A$ (AA), so $\frac{P B}{A B} = \frac{A B}{O B}$
$P B = \frac{A B ^{2}}{O B} = \frac{64}{6} = \frac{32}{3}$ cm

Also asked in: 2026 Standard 30/3/2, 2026 Standard 30/3/3

Q293 marksShort AnswerCircles CBSE 2025 · Basic 430/1/1

Prove that a rectangle circumscribing a circle is a square.

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Answer: Proved.

Let rectangle ABCD circumscribe a circle, touching AB, BC, CD, DA at P, Q, R, S.
Tangents from an external point are equal: AP = AS, BP = BQ, CR = CQ, DR = DS.
Adding: AB + CD = AD + BC.
In a rectangle AB = CD and BC = AD, so 2AB = 2BC, i.e. AB = BC.
A rectangle with adjacent sides equal is a square.

Also asked in: 2025 Basic 430/1/2, 2025 Basic 430/1/3

Q293 marksShort AnswerCircles CBSE 2025 · Basic 430/2/1

In the given figure, TP and TQ are tangents at points P and Q of the circle respectively. If reflex $∠ P O Q = 25 0^{\circ}$ , find the measure of each angle of quadrilateral POQT.

Diagram for CBSE 2025 Class 10 Maths question 29

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Answer:

∠ P O Q = 11 0^{\circ}

∠ O P T = 9 0^{\circ}

∠ O QT = 9 0^{\circ}

∠ P T Q = 7 0^{\circ}

$∠ P O Q = 36 0^{\circ} - 25 0^{\circ} = 11 0^{\circ}$ .
A tangent is perpendicular to the radius at the point of contact, so $∠ O P T = ∠ O QT = 9 0^{\circ}$ .
Sum of angles of quadrilateral POQT is $36 0^{\circ}$ : $∠ P T Q = 36 0^{\circ} - 11 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} = 7 0^{\circ}$ .

Also asked in: 2025 Basic 430/2/2, 2025 Basic 430/2/3

Q313 marksShort AnswerCircles CBSE 2025 · Basic 430/3/1

In the given figure, a circle is inscribed in a quadrilateral ABCD which touches the sides AB, BC, CD and DA at P, Q, R and S respectively. Prove that $∠ A O B + ∠ C O D = 18 0^{\circ}$ .

Diagram for CBSE 2025 Class 10 Maths question 31

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Answer: Proved.

Join OP, OQ, OR, OS. In $△ O A P$ and $△ O A S$ : $A P = A S$ (tangents from A), $O P = O S$ (radii), OA common, so they are congruent (SSS) and $∠ O A P = ∠ O A S$ .
Similarly OB, OC and OD bisect $∠ B$ , $∠ C$ and $∠ D$ . Let $∠ O A B = a$ , $∠ O B A = b$ , $∠ O C D = c$ , $∠ O D C = d$ .
Then $2 a + 2 b + 2 c + 2 d = ∠ A + ∠ B + ∠ C + ∠ D = 36 0^{\circ}$ , so $a + b + c + d = 18 0^{\circ}$ .
In $△ A O B$ : $∠ A O B = 18 0^{\circ} - (a + b)$ . In $△ C O D$ : $∠ C O D = 18 0^{\circ} - (c + d)$ .
$∠ A O B + ∠ C O D = 36 0^{\circ} - (a + b + c + d) = 36 0^{\circ} - 18 0^{\circ} = 18 0^{\circ}$ .

Also asked in: 2025 Basic 430/3/2, 2025 Basic 430/3/3

Q263 marksShort AnswerCircles CBSE 2025 · Standard 30/1/1

In the given figure, $O$ is the centre of the circle and $B C D$ is tangent to it at $C$ . Prove that $∠ B A C + ∠ A C D = 9 0^{\circ}$ .

Diagram for CBSE 2025 Class 10 Maths question 26

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Answer: Proved.

Join $O C$ . Since $B C D$ is tangent at $C$ , $O C ⊥ B D$ , so $∠ O C D = 9 0^{\circ}$ .
$∠ O C D = ∠ O C A + ∠ A C D$ , so $∠ O C A + ∠ A C D = 9 0^{\circ}$ ... (1)
In $△ O A C$ , $O A = O C$ (radii), so $∠ O A C = ∠ O C A$ .
$B, P, O, A$ are collinear, so $∠ B A C = ∠ O A C = ∠ O C A$ .
Substituting in (1): $∠ B A C + ∠ A C D = 9 0^{\circ}$ .

Also asked in: 2025 Standard 30/1/2, 2025 Standard 30/1/3

Q313 marksShort AnswerCircles CBSE 2025 · Standard 30/3/1

In the given figure, PC is a tangent to the circle at C. AOB is the diameter which when extended meets the tangent at P. Find $∠ C B A$ and $∠ B C O$ , if $∠ P C A = 11 0^{\circ}$ .

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Answer:

∠ C B A = 7 0^{\circ}

∠ B C O = 7 0^{\circ}

$∠ O C P = 9 0^{\circ}$ (radius $⊥$ tangent), so $∠ O C A = 11 0^{\circ} - 9 0^{\circ} = 2 0^{\circ}$ .
$O A = O C$ , so $∠ O A C = ∠ O C A = 2 0^{\circ}$ .
$∠ A C B = 9 0^{\circ}$ (angle in a semicircle).
In $△ A B C$ , $∠ C B A = 18 0^{\circ} - 9 0^{\circ} - 2 0^{\circ} = 7 0^{\circ}$ .
$O B = O C$ , so $∠ B C O = ∠ O B C = 7 0^{\circ}$ .

Also asked in: 2025 Standard 30/3/2

Q273 marksShort AnswerCircles CBSE 2025 · Standard 30/3/3

In the given figure, PB is a tangent to the circle with centre O at B. AB is a chord of the circle of length 24 cm and at a distance of 5 cm from the centre of the circle. If the length PB of the tangent is 20 cm, find the length of OP.

Diagram for CBSE 2025 Class 10 Maths question 27

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Answer:

O P = 569

cm (

\approx 23.85

cm)

OM $⊥$ AB, so M is the mid-point of AB: $M B = 12$ cm, $O M = 5$ cm.
In right $△ O M B$ , $O B = 1 2^{2} + 5^{2} = 169 = 13$ cm (radius).
$O B ⊥ P B$ (radius $⊥$ tangent), so in right $△ O B P$ , $O P^{2} = O B^{2} + P B^{2} = 169 + 400 = 569$ .
$O P = 569$ cm $\approx 23.85$ cm.

Q313 marksShort AnswerCircles CBSE 2025 · Standard 30/4/1

Rectangle ABCD circumscribes the circle of radius 10 cm. Prove that ABCD is a square. Hence, find the perimeter of ABCD.

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Answer: ABCD is a square (proved); perimeter

= 80

Let the sides AB, BC, CD, DA touch the circle at P, Q, R, S.
Tangents from an external point are equal: $A P = A S$ , $B P = B Q$ , $C R = C Q$ , $D R = D S$
Adding: $A B + C D = A D + B C$
In a rectangle $A B = C D$ , $A D = B C$ , so $2 A B = 2 B C$ , i.e. $A B = B C$ . A rectangle with adjacent sides equal is a square.
Side of square $=$ diameter $= 20$ cm, so perimeter $= 4 \times 20 = 80$ cm

Also asked in: 2025 Standard 30/4/2, 2025 Standard 30/4/3

Q313 marksShort AnswerCircles CBSE 2025 · Standard 30/5/1

In the adjoining figure, $X Y$ and $X^{'} Y^{'}$ are parallel tangents to a circle with centre O. Another tangent AB touches the circle at C intersecting $X Y$ at A and $X^{'} Y^{'}$ at B. Prove that AB subtends right angle at the centre of the circle; or $∠ A O B = 9 0^{\circ}$ .

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Answer: Proved.

Let XY touch the circle at P and X $^{'}$ Y $^{'}$ at Q. Join OC.
In $△ O P A$ and $△ O C A$ : $O P = O C$ (radii), $A P = A C$ (tangents from A), OA common. So the triangles are congruent (SSS) and $∠ O A P = ∠ O A C$ , i.e. $∠ O A B = \frac{1}{2} ∠ P A B$ .
Similarly $△ O QB ≅ △ O C B$ , so $∠ O B A = \frac{1}{2} ∠ QB A$ .
$X Y ∥ X^{'} Y^{'}$ and AB is a transversal, so $∠ P A B + ∠ QB A = 18 0^{\circ}$ (co-interior angles).
$∠ O A B + ∠ O B A = 9 0^{\circ}$
In $△ A O B$ : $∠ A O B = 18 0^{\circ} - 9 0^{\circ} = 9 0^{\circ}$ . Proved.

Also asked in: 2025 Standard 30/5/2, 2025 Standard 30/5/3

Q313 marksShort AnswerCircles CBSE 2025 · Standard 30/6/1

In the adjoining figure, TP and TQ are tangents drawn to a circle with centre O. If $∠ O P Q = 1 5^{\circ}$ and $∠ P T Q = θ$ , then find the value of $sin 2 θ$ .

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Answer:

sin 2 θ = \frac{3}{2}

$∠ O P T = 9 0^{\circ}$ , so $∠ T P Q = 9 0^{\circ} - 1 5^{\circ} = 7 5^{\circ}$
$T P = T Q$ , so $∠ T QP = ∠ T P Q = 7 5^{\circ}$
$θ = 18 0^{\circ} - 7 5^{\circ} - 7 5^{\circ} = 3 0^{\circ}$
$sin 2 θ = sin 6 0^{\circ} = \frac{3}{2}$

Also asked in: 2025 Standard 30/6/2, 2025 Standard 30/6/3

Q273 marksShort AnswerCircles CBSE 2024 · Basic 430/1/2

In two concentric circles, the radii are OA = r cm and OQ = 6 cm, as shown in the figure. Chord CD of larger circle is a tangent to smaller circle at Q. PA is tangent to larger circle. If PA = 16 cm and OP = 20 cm, find the length CD.

Diagram for CBSE 2024 Class 10 Maths question 27

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Answer: CD =

123

$O A ⊥ P A$ (radius is perpendicular to tangent), so $O A^{2} = O P^{2} - P A^{2} = 400 - 256 = 144$ and $r = 12$ cm.
$O Q ⊥ C D$ , and the perpendicular from the centre bisects the chord, so $C Q = Q D$ .
In $△ O Q D$ : $Q D = O D^{2} - O Q^{2} = 144 - 36 = 108 = 63$ cm.
$C D = 2 \times 63 = 123$ cm.

Also asked in: 2024 Basic 430/1/3

Q27 (OR) (OR)3 marksShort AnswerCircles CBSE 2024 · Basic 430/1/2

In given figure, two tangents PT and QT are drawn to a circle with centre O from an external point T. Prove that $∠ P T Q = 2∠ O P Q$ .

Diagram for CBSE 2024 Class 10 Maths question 27 (OR)

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Answer: Proved.

Let $∠ P T Q = θ$ .
TP = TQ (tangents from an external point), so $∠ T P Q = ∠ T QP = \frac{18 0 ^{\circ} - θ}{2} = 9 0^{\circ} - \frac{θ}{2}$ .
$∠ O P T = 9 0^{\circ}$ (radius is perpendicular to tangent).
$∠ O P Q = 9 0^{\circ} - ∠ T P Q = 9 0^{\circ} - (9 0^{\circ} - \frac{θ}{2}) = \frac{θ}{2}$ .
Hence $∠ P T Q = 2∠ O P Q$ .

Also asked in: 2024 Basic 430/1/3

Q273 marksShort AnswerCircles CBSE 2024 · Basic 430/2/1

In two concentric circles, a chord of length 24 cm of larger circle touches the smaller circle, whose radius is 5 cm. Find the radius of the larger circle.

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Answer: 13 cm

Let the chord AB touch the smaller circle at P. Then $O P ⊥ A B$ and OP = 5 cm.
The perpendicular from the centre bisects the chord, so AP = 12 cm.
In right $△ O P A$ , $O A^{2} = O P^{2} + A P^{2} = 25 + 144 = 169$
OA = 13 cm, so the radius of the larger circle is 13 cm.

Also asked in: 2024 Basic 430/2/2, 2024 Basic 430/2/3

Q273 marksShort AnswerCircles CBSE 2024 · Basic 430/4/1

Prove that opposite sides of a quadrilateral circumscribing a circle subtends supplementary angles at the centre of the circle.

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Answer: Proved.

Let ABCD circumscribe a circle with centre O, touching AB, BC, CD, DA at P, Q, R, S. Join OA, OB, OC, OD, OP, OQ, OR, OS.
$△ O A P ≅ △ O A S$ (RHS: $O P = O S$ , OA common, right angles at P and S), so $∠ A O P = ∠ A O S$ .
Similarly $∠ B O P = ∠ B O Q$ , $∠ C O Q = ∠ C O R$ , $∠ D O R = ∠ D O S$ .
The eight angles at O add to $36 0^{\circ}$ : $2 (∠ A O P + ∠ B O P + ∠ C O R + ∠ D O R) = 36 0^{\circ}$ .
So $(∠ A O P + ∠ B O P) + (∠ C O R + ∠ D O R) = 18 0^{\circ}$ , i.e. $∠ A O B + ∠ C O D = 18 0^{\circ}$ .
Similarly $∠ B O C + ∠ A O D = 18 0^{\circ}$ . Hence opposite sides subtend supplementary angles at O.

Also asked in: 2024 Basic 430/4/2, 2024 Basic 430/4/3

Q27 (OR) (OR)3 marksShort AnswerCircles CBSE 2024 · Basic 430/4/1

If O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of $4 0^{\circ}$ with PQ, then find the measure of $∠ P O Q$ .

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Answer:

∠ P O Q = 8 0^{\circ}

OP is a radius and PR is a tangent at P, so $∠ O P R = 9 0^{\circ}$ .
$∠ O P Q = 9 0^{\circ} - 4 0^{\circ} = 5 0^{\circ}$ .
$O P = O Q$ (radii), so $∠ O QP = ∠ O P Q = 5 0^{\circ}$ .
$∠ P O Q = 18 0^{\circ} - 5 0^{\circ} - 5 0^{\circ} = 8 0^{\circ}$ .

Also asked in: 2024 Basic 430/4/2, 2024 Basic 430/4/3

Q29 (OR) (OR)3 marksShort AnswerCircles CBSE 2024 · Basic 430/5/1

Prove that the lengths of the tangents drawn from an external point to a circle are equal.

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Answer: Proved.

Let PA and PB be tangents from an external point P to a circle with centre O, touching it at A and B.
Join OA, OB and OP. $∠ O A P = ∠ O B P = 9 0^{\circ}$ (radius $⊥$ tangent).
In right triangles OAP and OBP: OA = OB (radii) and OP is common.
So $△ O A P ≅ △ O B P$ (RHS).
Hence PA = PB (CPCT).

Also asked in: 2024 Basic 430/5/2, 2024 Basic 430/5/3

Q303 marksShort AnswerCircles CBSE 2024 · Standard 30/1/1

In the given figure, AB is a diameter of the circle with centre O. AQ, BP and PQ are tangents to the circle. Prove that $∠ P O Q = 9 0^{\circ}$ .

Diagram for CBSE 2024 Class 10 Maths question 30

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Answer: Proved.

Let PQ touch the circle at C. Join OC.
In $△ O A Q$ and $△ O C Q$ : OA = OC (radii), QA = QC (tangents from Q), OQ common. So $△ O A Q ≅ △ O C Q$ (SSS) and $∠ A O Q = ∠ C O Q$ .
Similarly $△ O B P ≅ △ O C P$ , so $∠ B O P = ∠ C O P$ .
AB is a diameter, so $∠ A O C + ∠ C O B = 18 0^{\circ}$ , i.e. $2∠ C O Q + 2∠ C O P = 18 0^{\circ}$ .
So $∠ C O Q + ∠ C O P = 9 0^{\circ}$ , i.e. $∠ P O Q = 9 0^{\circ}$ .

Also asked in: 2024 Standard 30/1/2, 2024 Standard 30/1/3

Q30 (OR) (OR)3 marksShort AnswerCircles CBSE 2024 · Standard 30/1/1

A circle with centre O and radius 8 cm is inscribed in a quadrilateral ABCD in which P, Q, R, S are the points of contact as shown. If AD is perpendicular to DC, BC = 30 cm and BS = 24 cm, then find the length DC.

Diagram for CBSE 2024 Class 10 Maths question 30 (OR)

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Answer: DC = 14 cm

Tangents from an external point are equal: BR = BS = 24 cm.
CR = BC $-$ BR = 30 $-$ 24 = 6 cm, so CQ = CR = 6 cm.
In quadrilateral OPDQ, $∠ D = 9 0^{\circ}$ and $∠ O P D = ∠ O Q D = 9 0^{\circ}$ (radius $⊥$ tangent), and OP = OQ = 8 cm, so OPDQ is a square.
So DQ = 8 cm.
DC = DQ + QC = 8 + 6 = 14 cm.

Also asked in: 2024 Standard 30/1/2, 2024 Standard 30/1/3

Q283 marksShort AnswerCircles CBSE 2024 · Standard 30/2/1

In the given figure, PQ is tangent to a circle centred at O and $∠ B A Q = 3 0^{\circ}$ ; show that BP = BQ.

Diagram for CBSE 2024 Class 10 Maths question 28

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Answer: Proved.

Join OQ. OA = OQ (radii), so $∠ O Q A = ∠ O A Q = 3 0^{\circ}$ .
AB is a diameter, so $∠ A QB = 9 0^{\circ}$ . Hence $∠ O QB = 9 0^{\circ} - 3 0^{\circ} = 6 0^{\circ}$ .
OQ $⊥$ PQ (tangent $⊥$ radius), so $∠ B QP = 9 0^{\circ} - 6 0^{\circ} = 3 0^{\circ}$ .
In $△ A QB$ , $∠ A B Q = 18 0^{\circ} - 9 0^{\circ} - 3 0^{\circ} = 6 0^{\circ}$ .
$∠ A B Q$ is an exterior angle of $△ B QP$ , so $∠ B P Q = 6 0^{\circ} - 3 0^{\circ} = 3 0^{\circ}$ .
$∠ B QP = ∠ B P Q$ , so BP = BQ (sides opposite equal angles).

Also asked in: 2024 Standard 30/2/2, 2024 Standard 30/2/3

Q28 (OR) (OR)3 marksShort AnswerCircles CBSE 2024 · Standard 30/2/1

In the given figure, AB, BC, CD and DA are tangents to the circle with centre O forming a quadrilateral ABCD.
Show that $∠ A O B + ∠ C O D = 18 0^{\circ}$

Diagram for CBSE 2024 Class 10 Maths question 28 (OR)

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Answer: Proved.

The two tangents from an external point are equally inclined to the line joining the point to the centre, so OA, OB, OC, OD bisect $∠ A$ , $∠ B$ , $∠ C$ , $∠ D$ .
In $△ A O B$ : $∠ A O B = 18 0^{\circ} - \frac{1}{2} (∠ A + ∠ B)$ .
In $△ C O D$ : $∠ C O D = 18 0^{\circ} - \frac{1}{2} (∠ C + ∠ D)$ .
Adding: $∠ A O B + ∠ C O D = 36 0^{\circ} - \frac{1}{2} (∠ A + ∠ B + ∠ C + ∠ D) = 36 0^{\circ} - \frac{1}{2} \times 36 0^{\circ} = 18 0^{\circ}$ .

Also asked in: 2024 Standard 30/2/2, 2024 Standard 30/2/3

Q313 marksShort AnswerCircles CBSE 2024 · Standard 30/3/1

Prove that the tangents drawn at the end points of a chord of a circle makes equal angles with the chord.

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Answer: Proved.

Let AB be a chord of a circle with centre O, and let the tangents at A and B meet at P.
Tangents drawn from an external point are equal, so PA = PB.
In $△ P A B$ , PA = PB, so $∠ P A B = ∠ P B A$ (angles opposite equal sides).
Hence the tangents make equal angles with the chord.
(If AB is a diameter the tangents are parallel and each makes $9 0^{\circ}$ with AB, so the angles are again equal.)

Also asked in: 2024 Standard 30/3/2

Q303 marksShort AnswerCircles CBSE 2024 · Standard 30/3/3

From an external point P, two tangents PA and PB are drawn to the circle with centre O. Prove that OP is the perpendicular bisector of chord AB.

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Answer: Proved.

Let OP meet AB at M.
In $△ P A M$ and $△ P B M$ : PA = PB (tangents from an external point are equal), $∠ A P M = ∠ B P M$ (OP bisects the angle between the tangents, since $△ O A P ≅ △ O B P$ by RHS), PM = PM (common).
So $△ P A M ≅ △ P B M$ (SAS), giving AM = BM and $∠ A M P = ∠ B M P$ .
$∠ A M P + ∠ B M P = 18 0^{\circ}$ , so each is $9 0^{\circ}$ .
Hence OP is the perpendicular bisector of chord AB.

Q283 marksShort AnswerCircles CBSE 2023 · Basic 430/1/1

Prove that the tangents drawn from an external point to a circle are equal in length.

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Answer: Proved.

Let PQ and PR be tangents from an external point P to a circle with centre O, touching it at Q and R. Join OQ, OR and OP.
$∠ O QP = ∠ O R P = 9 0^{\circ}$ (tangent $⊥$ radius at the point of contact).
In right triangles OQP and ORP: OQ = OR (radii) and OP = OP (common).
So $△ O QP ≅ △ O R P$ (RHS).
Hence PQ = PR (CPCT).

Also asked in: 2023 Basic 430/1/2, 2023 Basic 430/1/3

Q283 marksShort AnswerCircles CBSE 2023 · Basic 430/2/1

A quadrilateral ABCD is drawn to circumscribe a circle, as shown in the figure. Prove that AB + CD = AD + BC.

Diagram for CBSE 2023 Class 10 Maths question 28

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Answer: Proved.

Let the circle touch AB, BC, CD and DA at P, Q, R and S respectively.
Tangents drawn from an external point to a circle are equal:
AP = AS, BP = BQ, CR = CQ, DR = DS
Adding: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC

Also asked in: 2023 Basic 430/2/2, 2023 Basic 430/2/3

Q28 (OR) (OR)3 marksShort AnswerCircles CBSE 2023 · Basic 430/2/1

Prove that the parallelogram circumscribing a circle is a rhombus.

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Answer: Proved.

Let parallelogram ABCD touch the circle at P, Q, R, S on AB, BC, CD, DA.
Equal tangents from external points: AP = AS, BP = BQ, CR = CQ, DR = DS
Adding: AB + CD = AD + BC
In a parallelogram AB = CD and AD = BC, so 2AB = 2AD, i.e. AB = AD.
Thus AB = BC = CD = DA, so ABCD is a rhombus.

Also asked in: 2025 Standard 30/2/1, 2025 Standard 30/2/2, 2025 Standard 30/2/3, 2025 Basic 430/4/2, 2024 Standard 30/4/1, 2024 Standard 30/4/2, 2024 Standard 30/4/3, 2024 Standard 30/5/1, 2024 Standard 30/5/2, 2024 Standard 30/5/3, 2024 Basic 430/5/1, 2024 Basic 430/5/2, 2024 Basic 430/5/3, 2023 Basic 430/2/2, 2023 Basic 430/2/3

Q313 marksShort AnswerCircles CBSE 2023 · Basic 430/4/1

Prove that the lengths of tangents drawn from an external point to a circle are equal.

Show answer & solution

Answer: Proved.

Let PQ and PR be tangents from an external point P to a circle with centre O, touching it at Q and R. Join OQ, OR and OP.
$∠ O QP = ∠ O R P = 9 0^{\circ}$ (radius is perpendicular to tangent).
In right triangles OQP and ORP: OQ = OR (radii) and OP = OP (common).
So $△ O QP ≅ △ O R P$ (RHS).
Hence PQ = PR (CPCT).

Also asked in: 2026 Standard 30/2/1, 2026 Standard 30/2/3, 2023 Basic 430/4/2, 2023 Basic 430/4/3

Q31 (OR) (OR)3 marksShort AnswerCircles CBSE 2023 · Basic 430/4/1

Two concentric circles with centre O are of radii 3 cm and 5 cm. Find the length of chord AB of the larger circle which touches the smaller circle at P.

Diagram for CBSE 2023 Class 10 Maths question 31 (OR)

Show answer & solution

Answer: AB = 8 cm

OP is a radius of the smaller circle and AB touches it at P, so $O P ⊥ A B$ .
The perpendicular from the centre bisects the chord, so AP = PB.
In right $△ O P A$ : $A P = O A^{2} - O P^{2} = 25 - 9 = 4$ cm.
AB $= 2 \times 4 = 8$ cm.

Also asked in: 2023 Basic 430/4/2, 2023 Basic 430/4/3

Q313 marksShort AnswerCircles CBSE 2023 · Basic 430/5/1

ABC is an isosceles triangle with AB = AC, circumscribed about a circle. Prove that BC is bisected at E.

Diagram for CBSE 2023 Class 10 Maths question 31

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Answer: Proved.

The sides AB, BC, CA touch the circle at D, E, F.
Tangents from an external point are equal: AD = AF, BD = BE, CE = CF.
AB = AC gives AD + DB = AF + FC.
Since AD = AF, DB = FC, hence BE = CE.
So E is the mid-point of BC, i.e. BC is bisected at E.

Also asked in: 2023 Basic 430/5/2, 2023 Basic 430/5/3

Q293 marksShort AnswerCircles CBSE 2023 · Standard 30/1/1

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

Diagram for CBSE 2023 Class 10 Maths question 29

Show answer & solution

Answer: Proved.

Let $∠ P T Q = θ$ .
TP = TQ (tangents from an external point), so $∠ T P Q = ∠ T QP = \frac{18 0 ^{\circ} - θ}{2} = 9 0^{\circ} - \frac{θ}{2}$ .
OP $⊥$ TP, so $∠ O P T = 9 0^{\circ}$ .
$∠ O P Q = 9 0^{\circ} - ∠ T P Q = 9 0^{\circ} - (9 0^{\circ} - \frac{θ}{2}) = \frac{θ}{2}$ .
Hence $∠ P T Q = 2∠ O P Q$ .

Also asked in: 2026 Standard 30/2/1, 2026 Standard 30/2/3, 2026 Basic 430/4/2, 2023 Standard 30/1/2, 2023 Standard 30/1/3, 2023 Standard 30/6/1, 2023 Standard 30/6/2

Q29 (OR) (OR)3 marksShort AnswerCircles CBSE 2023 · Standard 30/1/1

In the given figure, a circle is inscribed in a quadrilateral ABCD in which $∠ B = 9 0^{\circ}$ . If AD = 17 cm, AB = 20 cm and DS = 3 cm, then find the radius of the circle.

Diagram for CBSE 2023 Class 10 Maths question 29 (OR)

Show answer & solution

Answer: 6 cm

Tangents from an external point are equal: DR = DS = 3 cm.
AR = AD – DR = 17 – 3 = 14 cm, so AQ = AR = 14 cm.
BQ = AB – AQ = 20 – 14 = 6 cm.
In OQBP, $∠ O QB = ∠ O P B = ∠ B = 9 0^{\circ}$ and BQ = BP, so OQBP is a square.
r = OQ = BQ = 6 cm.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q313 marksShort AnswerCircles CBSE 2023 · Standard 30/2/1

In the given figure, O is the centre of the circle and QPR is a tangent to it at P. Prove that $∠ Q A P + ∠ A P R = 9 0^{\circ}$ .

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Answer: Proved.

OP is a radius and QPR is the tangent at P, so $∠ O P R = 9 0^{\circ}$ .
OA = OP (radii), so $∠ O A P = ∠ O P A$ .
Q, B, O, A are collinear, so $∠ Q A P = ∠ O A P = ∠ O P A$ .
$∠ A P R = ∠ O P R - ∠ O P A = 9 0^{\circ} - ∠ Q A P$
Hence $∠ Q A P + ∠ A P R = 9 0^{\circ}$ .

Also asked in: 2023 Standard 30/2/2, 2023 Standard 30/2/3

Q283 marksShort AnswerCircles CBSE 2023 · Standard 30/4/1

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

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Answer: Proved.

Let PA and PB be tangents from external point P to a circle with centre O, touching it at A and B.
A radius is perpendicular to the tangent at the point of contact, so $∠ O A P = ∠ O B P = 9 0^{\circ}$ .
In quadrilateral OAPB, the angles add up to $36 0^{\circ}$ :
$∠ A P B + ∠ A O B + 9 0^{\circ} + 9 0^{\circ} = 36 0^{\circ}$
$∠ A P B + ∠ A O B = 18 0^{\circ}$
Hence the angle between the tangents is supplementary to $∠ A O B$ .

Also asked in: 2025 Standard 30/2/1, 2025 Standard 30/2/2, 2025 Standard 30/2/3, 2025 Basic 430/4/1, 2025 Basic 430/4/2, 2025 Basic 430/4/3, 2024 Basic 430/2/1, 2024 Basic 430/2/2, 2024 Basic 430/2/3, 2023 Standard 30/4/2

Q303 marksShort AnswerCircles CBSE 2023 · Standard 30/4/1

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

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Answer: 8 cm

Let O be the centre and AB the chord of the larger circle touching the smaller circle at P.
OP $⊥$ AB (radius $⊥$ tangent), and the perpendicular from the centre bisects the chord, so AP = PB.
In right $△ O P A$ : $A P = O A^{2} - O P^{2} = 25 - 9 = 4$ cm
$A B = 2 \times 4 = 8$ cm

Also asked in: 2023 Standard 30/4/2, 2023 Standard 30/4/3

Q273 marksShort AnswerCircles CBSE 2023 · Standard 30/4/3

From an external point, two tangents are drawn to a circle. Prove that the line joining the external point to the centre of the circle bisects the angle between the two tangents.

Show answer & solution

Answer: Proved.

Let PA and PB be tangents from external point P to a circle with centre O, touching it at A and B. Join OA, OB and OP.
In $△ O A P$ and $△ O B P$ :
OA = OB (radii); $∠ O A P = ∠ O B P = 9 0^{\circ}$ (radius $⊥$ tangent); OP is common.
So $△ O A P ≅ △ O B P$ (RHS congruence).
Hence $∠ O P A = ∠ O P B$ (CPCT), i.e. OP bisects $∠ A P B$ .

Q83 marksShort AnswerCircles CBSE 2022 · Basic 430/1/1Not in current syllabus

Draw a circle of radius 4 cm. Construct a pair of tangents to the circle from a point 6 cm away from its centre.

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Answer: Construction (see steps).

Draw a circle with centre O and radius 4 cm; mark P with OP = 6 cm
Bisect OP; let M be its midpoint
With M as centre and radius MO, draw a circle cutting the given circle at A and B
Join PA and PB; these are the required tangents (each about 4.5 cm long)

Also asked in: 2022 Basic 430/1/2

Q93 marksShort AnswerCircles CBSE 2022 · Basic 430/1/3Not in current syllabus

Draw a circle of radius 3 cm. Take a point P at a distance of 8 cm from the centre of the circle. Construct a pair of tangents from the point P to the circle.

Show answer & solution

Answer: Construction (see steps).

Draw a circle with centre O and radius 3 cm; mark P with OP = 8 cm
Bisect OP; let M be its midpoint
With M as centre and radius MO, draw a circle cutting the given circle at A and B
Join PA and PB; these are the required tangents (each $55 \approx 7.4$ cm long)

Q73 marksShort AnswerCircles CBSE 2022 · Basic 430/2/1Not in current syllabus

Draw a circle of radius 2.5 cm. Construct a pair of tangents from a point P at a distance of 6 cm from the centre of the circle.

Show answer & solution

Answer: Construction (pair of tangents from P drawn; each tangent ≈ 5.45 cm).

Draw a circle with centre O and radius 2.5 cm; mark P with OP = 6 cm.
Bisect OP; let M be its midpoint.
With M as centre and MO as radius draw a circle cutting the given circle at Q and R.
Join PQ and PR; these are the required tangents.
Justification: $∠ O QP = 9 0^{\circ}$ (angle in a semicircle), so PQ $⊥$ OQ; similarly PR.
Length of each tangent $= 6^{2} - 2. 5^{2} = 29.75 \approx 5.45$ cm.

Q103 marksShort AnswerCircles CBSE 2022 · Basic 430/2/3Not in current syllabus

Draw two concentric circles of radii 2 cm and 5 cm. From a point P on outer circle, construct a pair of tangents to the inner circle.

Show answer & solution

Answer: Construction (pair of tangents drawn; each tangent

= 21 \approx 4.58

cm).

Draw concentric circles with centre O and radii 2 cm and 5 cm; take P on the outer circle.
Bisect OP; let M be its midpoint.
With centre M and radius MO draw a circle cutting the inner circle at Q and R.
Join PQ and PR; these are the required tangents.
Justification: $∠ O QP = 9 0^{\circ}$ (angle in a semicircle), so PQ $⊥$ OQ; similarly PR.
Length of each tangent $= 5^{2} - 2^{2} = 21 \approx 4.58$ cm.

Q73 marksShort AnswerCircles CBSE 2022 · Basic 430/3/1Not in current syllabus

Draw a circle of radius 2.5 cm. From a point P lying outside the circle at a distance of 6 cm from the centre of the circle, construct tangents PA and PB to the circle.

Show answer & solution

Answer: Construction (each tangent

= 6^{2} - 2. 5^{2} = 29.75 \approx 5.45

cm)

Draw a circle with centre O and radius 2.5 cm; mark P with OP = 6 cm.
Draw the perpendicular bisector of OP; let it meet OP at M.
With M as centre and MO as radius, draw a circle cutting the given circle at A and B.
Join PA and PB; these are the required tangents.
Justification: $∠ O A P = ∠ O B P = 9 0^{\circ}$ (angles in a semicircle), so PA and PB are tangents. Each measures $36 - 6.25 \approx 5.45$ cm.

Also asked in: 2022 Basic 430/3/3

Q93 marksShort AnswerCircles CBSE 2022 · Basic 430/3/2Not in current syllabus

Draw two concentric circles of radii 5 cm and 2 cm. From a point P on the outer circle, construct a pair of tangents to the inner circle.

Show answer & solution

Answer: Construction (each tangent

= 5^{2} - 2^{2} = 21 \approx 4.58

cm)

Draw concentric circles with centre O and radii 2 cm and 5 cm; take P on the outer circle.
Draw the perpendicular bisector of OP; let it meet OP at M.
With M as centre and MO as radius, draw a circle cutting the inner circle at A and B.
Join PA and PB; these are the required tangents ( $∠ O A P = ∠ O B P = 9 0^{\circ}$ , angles in a semicircle).
Each tangent measures $25 - 4 = 21 \approx 4.58$ cm.

Q83 marksShort AnswerCircles CBSE 2022 · Standard 30/1/1Not in current syllabus

Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Construct tangents to the circle from these two points P and Q.

Show answer & solution

Answer: Construction: two tangents from each of P and Q; each tangent is

210 \approx 6.3

cm long.

Draw a circle with centre O and radius 3 cm. Draw a diameter and extend it both ways; mark P and Q on it with OP = OQ = 7 cm.
Bisect OP; let M be its midpoint. With M as centre and MO as radius, draw a circle cutting the given circle at A and B.
Join PA and PB: these are the tangents from P.
Repeat with the midpoint N of OQ to get points C and D; join QC and QD: these are the tangents from Q.
Justification: $∠ O A P = 9 0^{\circ}$ (angle in a semicircle), so PA $⊥$ OA and PA is a tangent.
Check: tangent length $= 7^{2} - 3^{2} = 40 \approx 6.3$ cm.

Q103 marksShort AnswerCircles CBSE 2022 · Standard 30/1/2Not in current syllabus

Construct a pair of tangents to a circle of radius 5 cm, which are inclined to each other at an angle of $6 0^{\circ}$ .

Show answer & solution

Answer: Construction: draw radii OA and OB with

∠ A O B = 12 0^{\circ}

; the perpendiculars to OA at A and to OB at B are the required tangents.

Draw a circle with centre O and radius 5 cm.
The angle between the tangents is $6 0^{\circ}$ , so the angle between the radii to the points of contact is $18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ .
Draw radius OA and radius OB with $∠ A O B = 12 0^{\circ}$ .
At A draw AP $⊥$ OA and at B draw BP $⊥$ OB; they meet at P.
PA and PB are the required tangents, since in quadrilateral OAPB, $∠ A P B = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 12 0^{\circ} = 6 0^{\circ}$ .

Also asked in: 2022 Standard 30/4/3

Q73 marksShort AnswerCircles CBSE 2022 · Standard 30/2/1Not in current syllabus

Draw two concentric circles of radii 2 cm and 5 cm. From a point on the outer circle, construct a pair of tangents to the inner circle.

Show answer & solution

Answer: Construction (each tangent is

21 \approx 4.6

cm long).

Draw circles of radii 2 cm and 5 cm with common centre O. Take a point P on the outer circle and join OP.
Bisect OP; let M be its midpoint.
With M as centre and MO as radius draw a circle cutting the inner circle at Q and R.
Join PQ and PR. These are the required tangents.
Justification: $∠ O QP = 9 0^{\circ}$ (angle in a semicircle), so PQ $⊥$ OQ and PQ is a tangent; similarly PR.
Length check: $P Q = 5^{2} - 2^{2} = 21 \approx 4.6$ cm.

Q103 marksShort AnswerCircles CBSE 2022 · Standard 30/2/2Not in current syllabus

Construct a pair of tangents to a circle of radius 3 cm which are inclined to each other at an angle of $6 0^{\circ}$ .

Show answer & solution

Answer: Construction (angle between the radii to the points of contact is

12 0^{\circ}

Draw a circle of radius 3 cm with centre O and draw a radius OA.
The angle between the radii to the points of contact is $18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ ; draw radius OB with $∠ A O B = 12 0^{\circ}$ .
At A and B draw lines perpendicular to OA and OB; let them meet at P.
PA and PB are the required tangents.
Justification: in quadrilateral OAPB, $∠ A = ∠ B = 9 0^{\circ}$ and $∠ O = 12 0^{\circ}$ , so $∠ A P B = 36 0^{\circ} - 30 0^{\circ} = 6 0^{\circ}$ .

Q103 marksShort AnswerCircles CBSE 2022 · Standard 30/2/3Not in current syllabus

Construct a pair of tangents to a circle of radius 4 cm from a point P lying outside the circle at a distance of 6 cm from the centre.

Show answer & solution

Answer: Construction (each tangent is

25 \approx 4.5

cm long).

Draw a circle of radius 4 cm with centre O and mark P with OP = 6 cm.
Bisect OP; let M be its midpoint.
With centre M and radius MO draw a circle cutting the given circle at Q and R.
Join PQ and PR; these are the required tangents.
Justification: $∠ O QP = 9 0^{\circ}$ (angle in a semicircle), so PQ $⊥$ OQ; similarly PR.
Length check: $P Q = 6^{2} - 4^{2} = 20 \approx 4.5$ cm.

Q7 (OR) (OR)3 marksShort AnswerCircles CBSE 2022 · Standard 30/3/1Not in current syllabus

Draw a circle of radius 3 cm. From a point P lying outside the circle at a distance of 6 cm from its centre, construct two tangents PA and PB to the circle.

Show answer & solution

Answer: Tangents PA and PB constructed; PA = PB =

33

cm ≈ 5.2 cm.

Draw a circle with centre O and radius 3 cm; mark P with OP = 6 cm.
Draw the perpendicular bisector of OP; let M be the midpoint of OP.
With M as centre and MO (= 3 cm) as radius, draw a circle cutting the given circle at A and B.
Join PA and PB; these are the required tangents.
Justification: $∠ O A P = 9 0^{\circ}$ (angle in a semicircle), so PA is a tangent; similarly PB.
Check: $P A = 6^{2} - 3^{2} = 27 = 33$ cm ≈ 5.2 cm.

Also asked in: 2022 Standard 30/3/2, 2022 Standard 30/3/3

Q73 marksShort AnswerCircles CBSE 2022 · Standard 30/4/1Not in current syllabus

Construct a pair of tangents to a circle of radius 4 cm which are inclined to each other at an angle of $6 0^{\circ}$ .

Show answer & solution

Answer: PA and PB drawn from the construction are the required tangents.

Draw a circle with centre O and radius 4 cm.
Draw any radius OA.
Since the angle between the tangents is $6 0^{\circ}$ , the angle between the radii to the points of contact is $18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ .
Draw radius OB such that $∠ A O B = 12 0^{\circ}$ .
At A and B draw lines perpendicular to OA and OB respectively; let them meet at P.
PA and PB are the required tangents, inclined to each other at $6 0^{\circ}$ .
Justification: in quadrilateral OAPB, $∠ O A P = ∠ O B P = 9 0^{\circ}$ and $∠ A O B = 12 0^{\circ}$ , so $∠ A P B = 36 0^{\circ} - 30 0^{\circ} = 6 0^{\circ}$ .

Also asked in: 2022 Standard 30/4/2

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