Circles: 5 marks Questions (CBSE Class 10)

10 different 5 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)

Q335 marksLong AnswerCircles CBSE 2026 · Standard 30/4/1

In the given figure, TP and TQ are tangents to a circle with centre M, touching another circle with centre N at A and B respectively. It is given that MQ = 13 cm, NB = 8 cm, BQ = 35 cm and TP = 80 cm.
(i) Name the quadrilateral MQBN. (1)
(ii) Is MN parallel to PA ? Justify your answer. (1)
(iii) Find length TB. (1)
(iv) Find length MN. (2)

Diagram for CBSE 2026 Class 10 Maths question 33

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Answer: (i) Trapezium (ii) No (iii) TB = 45 cm (iv) MN =

252

(i) $M Q ⊥ QT$ and $N B ⊥ QT$ , so $M Q ∥ N B$ ; MQBN is a trapezium (right trapezium).
(ii) $M P ⊥ P A$ and $N A ⊥ P A$ , so $M P ∥ N A$ . If also $M N ∥ P A$ , MPAN would be a parallelogram and MP = NA; but MP = 13 cm $\neq =$ NA = 8 cm. So MN is not parallel to PA.
(iii) Tangents from T: TQ = TP = 80 cm, so TB = TQ $-$ BQ = 80 $-$ 35 = 45 cm.
(iv) Draw $N L ⊥ M Q$ . Then LQ = NB = 8, ML = 13 $-$ 8 = 5 cm and NL = BQ = 35 cm.
$M N = 5^{2} + 3 5^{2} = 1250 = 252$ cm

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q355 marksLong AnswerCircles CBSE 2026 · Standard 30/5/1

PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

Diagram for CBSE 2026 Class 10 Maths question 35

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Answer: AB =

\frac{20}{3}

cm, PA =

\frac{26}{3}

$O Q ⊥ P Q$ , so $P Q = O P^{2} - O Q^{2} = 169 - 25 = 12$ cm.
C lies on OP and OC = 5 cm, so PC = 13 − 5 = 8 cm. $A B ⊥ O P$ at C.
Let AC = x. Tangents from A are equal, so AQ = AC = x and PA = 12 − x.
In right $△ A C P$ : $P A^{2} = A C^{2} + P C^{2}$
$(12 - x)^{2} = x^{2} + 64$ , so $144 - 24 x = 64$ and $x = \frac{10}{3}$ cm.
PA $= 12 - \frac{10}{3} = \frac{26}{3}$ cm
By symmetry BC = AC, so AB = 2x $= \frac{20}{3}$ cm.

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q335 marksLong AnswerCircles CBSE 2025 · Basic 430/6/1

The given figure shows a circle with centre O and radius 4 cm circumscribed by $△ A B C$ . BC touches the circle at D such that BD = 6 cm, DC = 10 cm. Find the length of AE.

Diagram for CBSE 2025 Class 10 Maths question 33

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Answer: AE

= \frac{64}{11}

cm (about 5.82 cm)

Tangents from an external point are equal: BF = BD = 6, CE = CD = 10, AE = AF = $x$ (say).
Sides: BC = 16, CA = $x + 10$ , AB = $x + 6$ ; semi-perimeter $s = x + 16$ .
Area $= \frac{1}{2} \times r \times$ perimeter $= 4 (x + 16)$ .
By Heron: Area $= s (s - a) (s - b) (s - c) = (x + 16) (x) (6) (10)$ .
So $16 (x + 16)^{2} = 60 x (x + 16)$ , i.e. $16 (x + 16) = 60 x$ .
$16 x + 256 = 60 x$ , so $x = \frac{256}{44} = \frac{64}{11}$ cm.

Also asked in: 2025 Basic 430/6/2, 2025 Basic 430/6/3

Q33 (OR) (OR)5 marksLong AnswerCircles CBSE 2025 · Basic 430/6/1

PA and PB are tangents drawn to a circle with centre O.
If $∠ A O B = 12 0^{\circ}$ and OA = 10 cm, then
(i) Find $∠ O P A$ . (1)
(ii) Find the perimeter of $△ O A P$ . (3)
(iii) Find the length of chord AB. (1)

Diagram for CBSE 2025 Class 10 Maths question 33 (OR)

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Answer: (i)

∠ O P A = 3 0^{\circ}

(ii)

(30 + 103)

cm (about 47.3 cm) (iii) AB

= 103

(i) $∠ O A P = ∠ O B P = 9 0^{\circ}$ , so $∠ A P B = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 12 0^{\circ} = 6 0^{\circ}$ . OP bisects $∠ A P B$ , so $∠ O P A = 3 0^{\circ}$ .
(ii) In right $△ O A P$ : $sin 3 0^{\circ} = \frac{O A}{O P}$ , so OP = 20 cm; $tan 3 0^{\circ} = \frac{O A}{A P}$ , so AP $= 103$ cm.
Perimeter $= 10 + 20 + 103 = (30 + 103)$ cm.
(iii) $△ A O B$ is isosceles with OA = OB = 10 and $∠ A O B = 12 0^{\circ}$ . The perpendicular from O to AB bisects AB and $∠ A O B$ .
Half of AB $= 10 sin 6 0^{\circ} = 53$ , so AB $= 103$ cm.

Also asked in: 2025 Basic 430/6/2, 2025 Basic 430/6/3

Q335 marksLong AnswerCircles CBSE 2024 · Basic 430/3/1

In the given figure, AB is chord of length 6 cm of a circle of radius 5 cm. The tangents at A and B intersect at a point P. Find the length of PB.

Diagram for CBSE 2024 Class 10 Maths question 33

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Answer:

P B = \frac{15}{4}

= 3.75

Let OP meet AB at M. Since $P A = P B$ and $O A = O B$ , OP is the perpendicular bisector of AB, so $A M = M B = 3$ cm.
In $△ O M B$ , $O M = 5^{2} - 3^{2} = 4$ cm.
Let $P M = y$ . In $△ P M B$ , $P B^{2} = y^{2} + 9$ .
$∠ O B P = 9 0^{\circ}$ , so $P B^{2} = O P^{2} - O B^{2} = (y + 4)^{2} - 25$ .
$y^{2} + 9 = y^{2} + 8 y + 16 - 25$ , so $8 y = 18$ and $y = \frac{9}{4}$ .
$P B^{2} = \frac{81}{16} + 9 = \frac{225}{16}$ , so $P B = \frac{15}{4} = 3.75$ cm.

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

Q33 (OR) (OR)5 marksLong AnswerCircles CBSE 2024 · Basic 430/3/1

Prove that the parallelogram circumscribing a circle is a rhombus. Also, find area of the rhombus, if radius of circle is 3 cm and length of one side of the rhombus is 10 cm.

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Answer: Proved; area of the rhombus = 60 cm

^{2}

Let parallelogram ABCD touch the circle at P, Q, R, S on AB, BC, CD, DA.
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 parallelogram $A B = C D$ and $A D = B C$ , so $2 A B = 2 B C$ , i.e. $A B = B C$ .
So all sides are equal and ABCD is a rhombus.
The distance between opposite sides of the rhombus equals the diameter of the circle $= 6$ cm.
Area = base $\times$ height $= 10 \times 6 = 60$ cm $^{2}$ .

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

Q335 marksLong AnswerCircles CBSE 2023 · Standard 30/5/1

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 10 cm and 8 cm respectively. Find the lengths of the sides AB and AC, if it is given that area $△ A B C = 90$ cm $^{2}$ .

Diagram for CBSE 2023 Class 10 Maths question 33

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Answer: AB = 14.5 cm, AC = 12.5 cm

Tangents from an external point are equal. Let the tangent length from A be x.
Then AB = x + 10, AC = x + 8, BC = 18 cm.
Area $△ A B C$ = area OBC + area OCA + area OAB $= \frac{1}{2} \times 4 \times (A B + B C + C A)$ .
$90 = 2 (x + 10 + 18 + x + 8) = 2 (2 x + 36)$ , so $2 x + 36 = 45$ , x = 4.5.
AB = 14.5 cm, AC = 12.5 cm.
Check: s = 22.5, area $= 22.5 \times 4.5 \times 8 \times 10 = 90$ cm $^{2}$ .

Also asked in: 2023 Standard 30/5/2, 2023 Standard 30/5/3

Q33 (OR) (OR)5 marksLong AnswerCircles CBSE 2023 · Standard 30/5/1

Two circles with centres O and O′ of radii 6 cm and 8 cm, respectively intersect at two points P and Q such that OP and O′P are tangents to the two circles. Find the length of the common chord PQ.

Diagram for CBSE 2023 Class 10 Maths question 33 (OR)

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Answer: PQ = 9.6 cm

O′P is tangent to the circle with centre O at P, so OP $⊥$ O′P; $∠ O P O^{'} = 9 0^{\circ}$ .
$O O^{'} = 6^{2} + 8^{2} = 10$ cm.
The common chord PQ is perpendicular to OO′ and bisected by it at A.
Area of $△ O P O^{'}$ : $\frac{1}{2} \times O P \times O^{'} P = \frac{1}{2} \times O O^{'} \times P A$ , so $P A = \frac{6 \times 8}{10} = 4.8$ cm.
PQ = 2PA = 9.6 cm.

Also asked in: 2023 Standard 30/5/2, 2023 Standard 30/5/3

Q32 (OR) (OR)5 marksLong AnswerCircles CBSE 2023 · Standard 30/6/1

A circle touches the side BC of a $△ A B C$ at a point P and touches AB and AC when produced at Q and R respectively. Show that AQ = $\frac{1}{2}$ (Perimeter of $△ A B C$ ).

Diagram for CBSE 2023 Class 10 Maths question 32 (OR)

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Answer: Proved.

Tangents from an external point are equal:
AQ = AR, BQ = BP, CP = CR.
Perimeter of $△ A B C$ = AB + BC + CA = AB + BP + PC + CA
= AB + BQ + CR + CA = AQ + AR = 2AQ.
So AQ = $\frac{1}{2}$ (Perimeter of $△ A B C$ ).

Also asked in: 2023 Standard 30/6/2

Q33 (OR) (OR)5 marksLong AnswerCircles CBSE 2023 · Standard 30/6/3

In the given figure, tangents PQ and PR are drawn to a circle such that $∠ R P Q = 3 0^{\circ}$ . A chord RS is drawn parallel to the tangent PQ. Find the measure of $∠ R QS$ .

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Answer:

∠ R QS = 3 0^{\circ}

PQ = PR (tangents from P), so $∠ P QR = ∠ P R Q = \frac{18 0 ^{\circ} - 3 0 ^{\circ}}{2} = 7 5^{\circ}$ .
RS $∥$ PQ, so $∠ S R Q = ∠ R QP = 7 5^{\circ}$ (alternate angles).
OQ $⊥$ PQ and RS $∥$ PQ, so OQ $⊥$ RS; the perpendicular from the centre bisects the chord RS, so Q is equidistant from R and S, i.e. QR = QS.
Hence $∠ QS R = ∠ QR S = 7 5^{\circ}$ .
$∠ R QS = 18 0^{\circ} - 7 5^{\circ} - 7 5^{\circ} = 3 0^{\circ}$

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