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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

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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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