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.
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.