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

Let the chord AB of the larger circle touch the smaller circle at point P.

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

Since OP ⊥ AB and OP is drawn from the centre to chord AB of the larger circle, it bisects AB.
So, AP = PB.

Using Pythagoras theorem in △OPB:
$$OB^2 = OP^2 + PB^2$$
$$5^2 = 4^2 + PB^2$$
$$PB^2 = 25 - 16 = 9$$
$$PB = 3 \text{ cm}$$

∴ Length of chord AB = 2 × PB = 6 cm

Source: Chapter 10, Section 10.2 & 10.3

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• The key property used: tangent ⊥ radius at point of contact (Theorem 10.1), which also means the perpendicular from the centre bisects the chord.
• Apply Pythagoras in the right triangle formed by the centre, the point of contact, and the endpoint of the chord.
• Always double the half-chord at the end — a common step students forget.
• This is directly similar to Exercise 10.2 Q.7 (radii 5 cm and 3 cm); here the smaller radius is 4 cm, giving PB = 3 cm and chord = 6 cm.