Mobius transformation for circles

Question

Let S1, S2, S3 be three nonintersecting circles whose centers do not belong to one line. Prove that there is a unique circle S orthogonal to S1, S2, S3.

I wanted to use a definition of the Mobius transformation to answer this and the theorem states:

Let S1 and S2 be two nonconcentric circles that do not intersect each other. Then there are points A and B which are symmetric with respect to S1 and with respect to S2.

My trouble is extending this definition to three circles instead of two, since I'm assuming their will be different symmetric points for each pair.

Answer 1

Use the concept of radical axis line of two circles. The Wikipedia article states:

In Euclidean geometry, the radical axis of two non-concentric circles is the set of points whose power with respect to the circles are equal.

Any two circles in general position have a radical axis in common. For any three circles $S_1, S_2, S_3$ in general position, the three pairs of circles correspond to three radical axis lines which intersect in a common point $P$. This point is the unique point whose power with respect to each of the circles are equal. The Wikipedia article also states:

The radical axis consists of all centers of circles, which intersect the given circles orthogonally.

Thus, the point $P$ is also the center of the unique circle $S$ which is orthogonal to the $S_1,S_2,S_3$ because it is also the center of the circle which passes through the centers of the three circles.