Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact. [Figure required: two concentric circles with common centre O; $C_1$ is the larger circle, $C_2$ the smaller. A chord AB of $C_1$ touches $C_2$ at point P.]

Answer: In two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

Step-by-step solution

Step 1: Define the setup

We are given two concentric circles, $C_1$ (the larger one) and $C_2$ (the smaller one), sharing a common center O. A chord AB of the larger circle $C_1$ is tangent to the smaller circle $C_2$ at point P. We need to prove that the point of contact P bisects the chord AB, meaning AP = BP.

Step 2: Establish tangency property

Since AB is tangent to the smaller circle $C_2$ at point P, and OP is the radius of $C_2$ passing through the point of contact, we can apply Theorem 10.1. This theorem states that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Hence, OP is perpendicular to AB.

Step 3: Apply perpendicular from center to chord property

Now consider AB as a chord of the larger circle $C_1$ . We have established that OP is perpendicular to AB. A fundamental property of circles states that a perpendicular drawn from the center of a circle to a chord bisects the chord. Therefore, the point P must bisect the chord AB, which means AP = BP. This completes the proof.

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