# Line containing centers of circles formed by intersection points of 3 circles passes through the radical center of the three circles

While messing around with the radical center of three circles, I found out the following thing:

Given three circles intersecting in the points $$M,N,O,P,Q,R$$ as in the picture, the line joining the circumcenters of the triangles $$\Delta MNO$$, $$\Delta PQR$$ passes through the radical center of the three circles.

Why is this true? I tried using power of a point wrt the two circles to show that the radical center lies on the line, but to no avail. Any ideas?

Let $$p$$ (assumed $$<0$$) be the common power of the radical center $$T$$ with respect to the 3 circles.

Consider inversion transform $$I$$ having $$T$$ as its center, and $$p$$ as its power.

$$I$$ exchanges $$M,N,O$$ and $$P,Q,R$$ ; thus, as an inversion sends circles onto circles, (circumscribed) circle $$MNO$$ is exchanged with (circumscribed) cercle $$PQR$$.

In such a case the center of inversion is aligned with the circle's centers. QED.

Remarks :

1) In fact, I isn't a "true" inversion, but as Coxeter and Greitzer define it, an anti-inversion, i.e., a composition $$I= S \circ I'$$, of a genuine inversion $$I'$$ (with same center $$T$$, opposite power $$-p>0$$) and a central symmetry $$S$$. (see p. 192 of their masterpiece "Geometry Revisited" http://www.aproged.pt/biblioteca/geometryrevisited_coxetergreitzer.pdf). No change in our reasoning because alignment is preserved by $$S$$.

2) "as circles are sent onto circles" : one could object that a circle could be exchanged with a straight line, but this happens only in the case where the circles pass through the origin, which is not the case here because of assumption $$p<0$$.

3) I have devoted a question to inversion (Ill-known/original/interesting investigations on/applications of inversion (the geometric transform)) some time ago. A certain number of properties of this transformation can be found there.