MathsCBSE Class 10 · Standard · Board exam 2026, set 30/5/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2026: question 35

Set in the CBSE Class 10 board exam of 2026, paper 30/5/1, section D, question 35, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35 · 5 marks · Long answerCircles

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.

Figure in the paper: A circle with centre O. P lies outside the circle on the right. Tangents PQ (touching at Q, upper point) and PR (touching at R, lower point) are drawn from P. A third tangent AB touches the circle at C, where C lies on segment OP; A lies on PQ and B lies on PR, so AB is a tangent segment crossing OP at C. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 30/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Start with right triangle OQP: the radius is perpendicular to the tangent at Q. Find PQ.

Hint 2

Tangents from A are equal, so AQ = AC. Call this x and write PA in terms of x.

Hint 3

Since OC ⟂ AB and C lies on OP, triangle ACP is right-angled at C. Apply Pythagoras with PC = OP − OC.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

In right $△ O QP$ , $∠ O QP = 9 0^{\circ}$ [tangent ⟂ radius].
So $P Q = O P^{2} - O Q^{2} = 169 - 25 = 144 = 12$ cm.
1 mark
$P C = O P - O C = 13 - 5 = 8$ cm.
Let $A C = A Q = x$ [tangents from an external point A are equal]. Then $P A = P Q - A Q = 12 - x$ .
1 mark
$O C ⊥ A B$ [tangent ⟂ radius] and C lies on OP, so $∠ A C P = 9 0^{\circ}$ .
In right $△ A C P$ : $P A^{2} = A C^{2} + P C^{2} \Rightarrow (12 - x)^{2} = x^{2} + 8^{2}$
1 mark
$144 - 24 x + x^{2} = x^{2} + 64 \Rightarrow 24 x = 80 \Rightarrow x = \frac{10}{3}$ cm
1 mark
In the same way, $B C = B R$ and $P B = 12 - B C$ ; with $P C = 8$ and $∠ B C P = 9 0^{\circ}$ this gives $B C = \frac{10}{3}$ cm.
So $A B = A C + C B = 2 \times \frac{10}{3} = \frac{20}{3}$ cm $\approx 6.67$ cm.
½ mark
$P A = 12 - \frac{10}{3} = \frac{26}{3}$ cm $\approx 8.67$ cm.
½ mark

Answer: AB = 20/3 cm (≈ 6.67 cm); PA = 26/3 cm (≈ 8.67 cm)

Where the marks are

Finds $P Q = 12$ cm using tangent ⟂ radius and Pythagoras (1 mark)
$P C = 8$ cm, and $P A = 12 - x$ with $A C = A Q = x$ (1 mark)
Uses $A B ⊥ O P$ to write $(12 - x)^{2} = x^{2} + 8^{2}$ (1 mark)
Solves to get $x = 10/3$ cm (1 mark)
$A B = 2 A C = 20/3$ cm (½ mark)
$P A = 26/3$ cm (½ mark)

Revise the chapter

Chapter: Circles (notes, formulas, where marks are lost and every past question)
Topic: Solving Problems Using Tangent Properties

Practise this chapterBoard-style questions from Circles, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 30/5/1 (Standard): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Standard 2026: question 35

Hints

Step-by-step solution

Revise the chapter

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