MathsCBSE Class 10 · Standard · Board exam 2023, set 30/5/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2023: question 33(b)

Set in the CBSE Class 10 board exam of 2023, paper 30/5/1, section D, question 33(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(b) · 5 marks · Long answerCircles

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.

Figure in the paper: Two intersecting circles with centres O (left) and O' (right). P is the upper intersection point and Q the lower one. Segment PQ is the common chord, and it meets line OO' at A. Segments OP and O'P are drawn, with tangent lines at P shown. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 30/5/1.

Hints

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

Hint 1

Since OP is a tangent to the other circle, what is the angle between OP and O'P?

Hint 2

The line joining the centres bisects the common chord at right angles; let OA = x and use Pythagoras in two triangles.

Hint 3

Equate the two expressions for AP², solve for x, then double AP.

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.

OP is tangent to the circle with centre O' at P, so $O P ⊥ O^{'} P$ [tangent ⟂ radius]. In right $△ O P O^{'}$ : $O O^{'} = 6^{2} + 8^{2} = 100 = 10$ cm.
½ mark
The line of centres OO' is the perpendicular bisector of the common chord PQ, so $P A ⊥ O O^{'}$ and $P Q = 2 A P$ . Let $O A = x$ cm, then $O^{'} A = (10 - x)$ cm.
In right $△ P A O$ : $A P^{2} = O P^{2} - O A^{2} = 36 - x^{2}$
½ mark
In right $△ P A O^{'}$ : $A P^{2} = O^{'} P^{2} - O^{'} A^{2} = 64 - (10 - x)^{2}$
½ mark
Equating: $36 - x^{2} = 64 - (10 - x)^{2} = 64 - 100 + 20 x - x^{2}$ $\Rightarrow 36 = - 36 + 20 x \Rightarrow 20 x = 72 \Rightarrow x = 3.6$ cm
2 marks
$A P^{2} = 36 - (3.6)^{2} = 36 - 12.96 = 23.04 \Rightarrow A P = 4.8$ cm
1 mark
Length of common chord $P Q = 2 \times A P = 2 \times 4.8 = 9.6$ cm.
½ mark

Answer: PQ = 9.6 cm

Where the marks are

Uses OP ⟂ O'P to get OO' = 10 cm (½ mark)
Writes AP² = 36 − x² in ΔPAO (½ mark)
Writes AP² = 64 − (10 − x)² in ΔPAO' (½ mark)
Equates and solves to get x = 3.6 (2 marks)
Finds AP = 4.8 cm (1 mark)
PQ = 2AP = 9.6 cm (½ mark)

Revise the chapter

Chapter: Circles (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Circles, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 30/5/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2023: question 33(b)

Hints

Step-by-step solution

Revise the chapter

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