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Question

# In the given figure, the chord AB of the larget of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.

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Solution

## Here is the solution for your question name are different to that of your question Let there is a circle having center O Let AB is the tangent to the smaller circle and chord to the larger circle. Let P is the point of contact. Now, draw a perpendicular OP to AB Now, since AB is the tangent to the smaller circle, So, ∠OPA = 90 Now, AB is the chord of the larger circle and OP is perpendicular to AB. Since the perpendicular drawn from the center of the circle to the chord bisect it. So, AP = PB Hence, in two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact

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