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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!)

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    Mathematicians are not responsible for memorizing the entire history of mathematics. I don't understand what you mean by $L$.2011-08-16
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    Welcome to the fray, Hauke! It's an interesting idea, one new to me.2011-08-16
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    Here's an image: http://imgur.com/ttjN4 The dotted circle is for @=120° (of course everything is drawn only approximate!)2011-08-16
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    L is not a line but 4 lines, right? (E.g. when C1-C3 become straight lines L is 4 points: centers of in/ex-circles.)2011-08-19
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    @Grigory: If we specify a certain direction for directed angles, as I assumed in my solution, then there is one line for each of the 4 configurations. (Fix the direction of $C$ and $C_1$, then there are $2 \times 2$ configurations for $C_2$ and $C_3$.)2011-12-31

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