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

Hey MATE!

The answer to your question is in above attachment.

Hope it helps

Hakuna Matata :))

Answer 2

Answer:

Step-by-step explanation:

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 a circle is perpendicular to the radius through the point of contact)

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,

Hence AP = BP