The Radical Axis Theorem states the following:
Given three distinct circles , the radical axes of the three circles taken pairwise concur at a point, called the radical center.
The vanilla method of proving this statement is not hard:
Suppose and has radical axis , and and has radical axis . Let . Then by the definition of radical axis, because and because . Naturally, this means so , and the proof is finished.
Amazingly, there is also a very nice alternate way to prove this statement by extending the diagram to three dimensions. This proof will only work if the three circles pairwise intersect, but I believe it's possible to extend the proof to any three circles using complex points. But let's focus on real situations for now.
Let the center of circle be for . In addition, let the intersection of circles and that is closer to be , and define similarly.
Note that by definition, , and cyclically similar expressions. Thus, if we imagine rotating about line and about line , then at some point, the image of and the image of about these two rotations will coincide at a point above the plane containing . In other words, is a tetrahedron with base . However, note that because we performed a rotation, and . Since and , , so we can rotate about line so that the image of coincides with .
But here's the punchline: if we track the projections of onto the plane containing as we rotate the triangles, it is clear to see that the locus is just a perpendicular from to line (where indices are taken mod 3)--but this is just another definition of the radical axis! Thus, the projection of onto the plane containing lies on the radical axis of each pair of circles, and the proof is complete.