Let (O) be the quadrilateral circumcircle and X the intersection of PQ and RS. Let $\sigma_1$ be an arbitrary circle passing through P, Q tangent at A to a circle $\sigma_2$ passing through R, S. PQ is the radical axis of $(O), \sigma_1$ and RS is the radical axis of $(O), \sigma_2$. Hence, X is the radical center of $(O), \sigma_1, \sigma_2$ independent of the circles $\sigma_1, \sigma_2$. The single common internal tangent (resp. the single common external tangent) at A of the circles $\sigma_1, \sigma_2$ is their radical axis, if they touch externally (resp. if they touch internally). Thus this common tangent passes through the radical center X. Since all possible circles $\sigma_1$ belong to the same pencil with the base points P, Q and since X is on the radical axis of this pencil, the tangent length XA is constant and equal to the square root of the power of the point X to any circle $\sigma_1$, i.e., $r = XA = \sqrt{XP \cdot XQ} = \sqrt{XR \cdot XS}$. Consequently, the locus of the tangency points A of the circles $\sigma_1, \sigma_2$ is a circle centered at X and with radius $r$.