## Radical Axis

#### 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.