Mathematics — Standard (041) — CBSE Class 10 board question

Let the two concentric circles have centre O, with radii R = 4 cm and r = 3 cm. Let AB be the chord of the larger circle that touches the smaller circle at point P.

Since AB is a tangent to the smaller circle, OP ⊥ AB (radius ⊥ tangent at point of contact).

Also, since OP ⊥ AB and OP is from the centre of the larger circle, P is the midpoint of AB (perpendicular from centre bisects the chord).

In right △OPB:
$$OB^2 = OP^2 + PB^2$$
$$4^2 = 3^2 + PB^2$$
$$PB^2 = 16 - 9 = 7$$
$$PB = \sqrt{7} \text{ cm}$$

Therefore, $AB = 2PB = 2\sqrt{7}$ cm.

Source: Chapter 10, Section 10.2 (Theorem 10.1) and Section 10.3

---

• The key insight is that the chord of the larger circle touches (is tangent to) the smaller circle, so the radius to the point of contact is perpendicular to the chord.
• This perpendicular from the centre also bisects the chord (standard chord property).
• Apply Pythagoras in the right triangle formed by the centre, the midpoint of the chord, and an endpoint.
• Examiners expect a labelled diagram (mentally or drawn), the Pythagoras step clearly shown, and the final answer as $2\sqrt{7}$ cm.