Hi Cheral!

In order to prove your question we will use one property. It can be stated as **Let PT be a tangent to the circle from an exterior point P and a secant to the circle through P intersects the circle at points A and B where T is a point on the circle, then PT**^{2} = PA.PB. First of all I will prove this and use it to prove your question.

Let PT be a tangent to the circle from an exterior point P and a secant to the circle through P intersects the circle at points A and B where T is a point on the circle

Using Pythagoras theorem for ∆OPT

OT^{2} + PT^{2} = OP^{2}

⇒*r* ^{2} + PT^{2} = *r* ^{2} + PA.PB [using (2)]

⇒PT^{2} = PA.PB … (3)

Now, I will use this result to prove your question.

The information provided by you is represented diagrammatically as

Here, the circles intersect at point X and Y. A is a point on the line joining the points X and Y. AM and AN are the tangents drawn to the circles

You need to prove AM = AN

Using (3), it can be said that

AM^{2} = AX.AY and AN^{2} = AX.AY

Thus, AM^{2} = AN^{2}

⇒AM = AN

Hence, proved

Hope! This will help you.

Cheers!