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; C1 is the larger circle, C2 the smaller. A chord AB of C1 touches C2 at point P.]
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Step-by-Step Solution
Step 1: Define the setup
We are given two concentric circles, C1 (the larger one) and C2 (the smaller one), sharing a common center O. A chord AB of the larger circle C1 is tangent to the smaller circle C2 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 C2 at point P, and OP is the radius of C2 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 C1. 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.