### Tangent as a Radical Axis

The applet below illustrates Problem 4 from the 1995 Asia Pacific Mathematical Olympiad:

 Let PQRS be a cyclic quadrilateral such that the segments PQ and RS are not parallel. Consider the set of circles through P and Q, and the set of circles through R and S. Determine the set I of points of tangency of circles of these two sets.

In the applet, T is an arbitrary point in the plane of the cyclic quadrilateral PQRS. The applet shows circumcircle of triangles PQT and RST. The configuration of interest is where the two circles touch at their common point. On such occasions T is renamed to I.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run? ### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

Let C be the circumcircle of the quadrilateral PQRS and C' and C'' the circumcircles of triangle PQT and RST, respectively. PQ is the radical axis of C and C'; RS is the radical axis of C and C''. Point J is the radical center of the three circles. It follows that the radical axis of C' and C'' passes through J. On the other hand, the radical axis of two touching circles is their common tangent through the point of tangency. Therefore, the tangent to the two circles C' and C'' at I passes through J.

According to the Intersecting Chords Theorem, IJ² = PJ×QJ (and also, of course, IJ² = RJ×SJ.) For the fixed P, Q (or R, S), the product on the right is fixed, implying that IJ is constant. This means that I is located on a circle centered at J of radius PJ×QJ.

When I lies on PQ, circle IPQ degenerates into a straight line, a circle of infinite radius. Similarly, when I lies on RS, circle IRS becomes a straight line. When I is inside C, C' and C'' touch externally; otherwise they touch internally.

(Elsewhere Vo Duc Dien offers a more detailed solution.)  