In Figure, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90° For Visually Impaired candidates: Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that ∠ APB = 2(∠ OAB)

MathsCBSE Class 10 · Standard · Sample paper, session 2025-26 · 3 marks · Short answer

CBSE Class 10 Mathematics Standard sample paper 2025-26: question 26

From the official CBSE sample question paper for 2025-26, section C, question 26, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q26 · 3 marks · Short answerCircles

In Figure, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that $∠ A O B = 9 0^{\circ}$

For Visually Impaired candidates: Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that $∠ A P B = 2 (∠ O A B)$

Figure in the paper: Circle with centre O between two parallel horizontal tangent lines XY (top, touching at P) and X′Y′ (bottom, touching at Q). A third tangent AB touches the circle at C, meeting XY at A and X′Y′ at B. OA, OB, OC, OP, OQ are joined. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

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

Hint 1

Join OC and compare triangle APO with triangle ACO.

Hint 2

OA bisects angle POC and OB bisects angle QOC.

Hint 3

P, O, Q lie on a straight line, so the angles at O on one side add to 180°.

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.

Join OC. In $△ A P O$ and $△ A C O$ : $A P = A C$ [tangents from external point A], $O P = O C$ [radii], $O A = O A$ [common]. So $△ A P O ≅ △ A C O$ [SSS], and hence $∠ P O A = ∠ C O A$ [CPCT].
In the same way, $△ B QO ≅ △ B C O$ , so $∠ QO B = ∠ C O B$ .
1 mark
$O P ⊥ X Y$ and $O Q ⊥ X^{'} Y^{'}$ [tangent ⟂ radius]. Since $X Y ∥ X^{'} Y^{'}$ , the points P, O, Q are collinear, so PQ is a diameter. Therefore
$∠ P O A + ∠ C O A + ∠ C O B + ∠ QO B = 18 0^{\circ}$ .
1 mark
Hence $2∠ C O A + 2∠ C O B = 18 0^{\circ}$ , so $∠ C O A + ∠ C O B = 9 0^{\circ}$ , i.e. $∠ A O B = 9 0^{\circ}$ . Hence proved.
1 mark
For Visually Impaired candidates (alternative): $P A = P B$ [tangents from external point], so $∠ P A B = ∠ P B A = x$ [angles opposite equal sides]. In $△ P A B$ , $∠ A P B = 18 0^{\circ} - 2 x$ … (i). (½ mark in VI version)
(VI) $∠ O A P = 9 0^{\circ}$ [radius ⟂ tangent], so $x + ∠ O A B = 9 0^{\circ}$ , giving $x = 9 0^{\circ} - ∠ O A B$ … (ii). (1 mark)
(VI) Putting (ii) in (i): $∠ A P B = 18 0^{\circ} - 2 (9 0^{\circ} - ∠ O A B) = 2∠ O A B$ . Hence proved. (1 + ½ marks)

Answer: ∠AOB = 90° (proved); VI version: ∠APB = 2∠OAB (proved)

Where the marks are

Proves $△ A P O ≅ △ A C O$ (and similarly for the triangles at B), getting equal angles (1 mark)
Uses PQ as a diameter so that the four angles at O add to 180° (1 mark)
Concludes $∠ A O B = 9 0^{\circ}$ (1 mark)
VI: equal tangents and base angles (½), angle APB = 180° − 2x (1), x = 90° − ∠OAB using radius ⟂ tangent (1), conclusion (½)

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 paperSample paper 2025-26 (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 sample paper 2025-26: question 26

Hints

Step-by-step solution

Revise the chapter

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