In two concentric circles, prove that a chord of larger circle which is tangent to smaller circle is bisected at the point of contact.

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Solution

Let $O$ be the common centre of two concentric circles, and let $A B$ be a chord of the larger circle touching the smaller circle at $P$ .

To prove: $A B$ is bisected at $P$ .

Join $O P$

Since $O P$ is the radius of the smaller circle and $A B$ is a tangent to this circle at a point $P$ .

$∴ O P ⊥ A B$

We know that the perpendicular drawn from the centre of a circle to any chord of the circle, bisects the chord.

So, $O P ⊥ A B$

$\Rightarrow A P = B P$

Hence, $A B$ is bisected at $P$ .

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Similar questions

Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle, is bisected at the point of contact.

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Using the above, do the following:

In fig., O is the center of the two concentric circles. AB is a chord of the larger circle touching the smaller circle at C. Prove that AC = BC

In Fig.4, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.

OR
In Fig.5, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.

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