If from any point on the common chord of two interesting circle,

Let the two circles intersect at points X and Y. XY is the common chord.

Suppose A is a point on the common chord and AM and AN be the tangents drawn from A to the circle.

We need to show that AM = AN.

In order to prove the above relation, following property will be used.

“Let PT be a tangent to the circle from an external point P and a secant to the circle through P intersects the circle at points A and B, then PT 2 = PA × PB”.

Now, AM is the tangent and AXY is a secant.

∴ AM2 = AX × AY ...(1)

AN is the tangent and AXY is a secant.

∴ AN2 = AX × AY ...(2)

From (1) and (2), we have

AM2 = AN2

∴ AM = AN