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Astonishingly, no mathematician ever could give a "Mr. Foobar invented this" whenever I came up with this construction, although it is very elementary.

Given are 3 circles C1,C2,C3 (avoid degenerate configurations for now). Let L be the geometric locus of the centers of all circles C which intersect C1, C2 and C3 under the same angle @ (which may be non-real - doesn't hurt!)

Clearly the radical center (@=90°) and the all-outer/inner Apollonius center (@=0/180°) lie on L, and some analytic geometry immediately shows L is a straight line.

Bonus Track (only if you have too much time): Calculate @ for the Gergonne point when L is the Soddy line of C1, C2, C3. A most surprising result awaits. (Purely geometric proof, anyone?)

Edit: (Added from comments)

Here's an image:

enter image description here

The dotted circle is for @=120° (of course everything is drawn only approximate!)

1 Answers 1

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I have no idea what the Soddy line is, but I think I have solved the first part:

(I only consider the cases where $C_1$, $C_2$, $C_3$ are pair-wise distinct)
Let the centre of $C$ be $O$ and the centres of $C_1$, $C_2$, $C_3$ be $O_1$, $O_2$, $O_3$
If $C_1$, $C_2$, $C_3$ have the same size,

    L is not a straight line but a point because the angle at the intersection of $C$ and $C_1$ is monotonic in $\overline{OO_1}$
    Thus $\overline{OO_1} = \overline{OO_2} = \overline{OO_3}$ and $O$ is unique

If $C_1$, $C_2$, $C_3$ do not all have the same size,

    I think no two Apollonius circles can be concentric
    Thus there is a point at which inversion maps two corresponding ones to concentric circles
    In that case $C_1$, $C_2$, $C_3$ map to circles of the same size, so we are back to the earlier case!
    $C$ must then map to a circle with centre $P$ uniquely defined by the images of $C_1$, $C_2$, $C_3$
    Thus the centre of $C$ must lie on the line uniquely defined by the inversion centre and $P$

(QED)

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    I forgot to mention that inversion preserves angles at intersections.2011-12-30