Question

In Fig. 5, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.

Answer 1

Given: Two concentric circles C₁ and C₂ with centre O, and AB is the chord of C₁ touching C₂ at C.

To prove: AC = CB

Construction: Join OC.

Proof: AB is the chord of C₁ touching C₂ at C, then AB is the tangent to C₂ at C with OC as its radius.

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OC ⊥ AB

Considering, AB as the chord of the circle C₁. So, OC ⊥ AB.

∴ OC is the bisector of the chord AB.

Hence, AC = CB (Perpendicular from the centre to the chord bisects the chord).