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Problem

Two circles, centred at A and B, intersect at X and Y. From a point, P, on the common secant through XY, tangents are drawn to each circle at S and T.

The locus of points, P, for which PT = PS is called the radical axis.

Prove that the common secant is the radical axis.

Solution

Consider the diagram below.

We note that the size of the two angles marked a are equal because of the alternate segment theorem (from chord TY); similarly with the size of the angles marked b (chord TX).

As b+c and d are complementary angles to a (with the straight line), it follows that d = b+c.

Therefore triangle PTY is similar to triangle PXY.

PT/PY = PX/PT, which gives, (PT)² = PX PY.

As a consequence of this result we can see in the original diagram that
(PT)² = PA PB and (PS)² = PX PY, hence PT = PS.

Prove that the radical axis for two non-intersecting circles is a line perpendicular to their centres.