Question

Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

Answer 1

We are given two concentric circles C₁ and C₂ with centre O and a chord AB of the larger circle C₁ which touches the smaller circle C₂ at the point P (see Fig. 10.8). 

We need to prove that AP = BP.

Let us join OP. Then, AB is a tangent to C₂ at P and OP is its radius. Therefore, 

OP⟂AB (Since Tangent at any point of circle is perpendicular to the radius through point of cntact)

Now AB is a chord of the circle C₁ and OP⟂AB. Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord,

i.e., AP = BP