TITLE: Is this concept of circle geometry known?
QUESTION [6 upvotes]: 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:  

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

REPLY [2 votes]: 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)