If a circle is mapped into another circle by inversion, then the centers of the two circles and the center of inversion must be collinear. This means for our problem that points P and Q must lie on line AB. Let O be a center of inversion on the line AB and at a distance x from A (x is positive if point A lies between points O and B). If k is the radius of the circle of inversion and O is to be P or Q, then
k2 k2 k2 k2
----- + ----- ------- + -------
x-a x+a x+c-b x+c+b
--------------- = -------------------
2 2
or
cx2 + (a2 - b2 + c2)x + a2c = 0
Solving for x,
-(a2 - b2 + c2) ± √([a2 - b2 + c2]2 - 4a2c2)
x = -------------------------------------------
2c
For there to be two real roots,
[a2 - b2 + c2]2 - 4a2c2 > 0
Case I - Circle A contains circle B: a > b+c.
[a2 - b2 + c2]2 - 4a2c2 = (a+b+c)(a+b-c)(a-b+c)(a-b-c) > 0
Case II - Circle B contains circle A: b > a+c.
[a2 - b2 + c2]2 - 4a2c2 = (b+c+a)(b+c-a)(b-c+a)(b-c-a) > 0
Case III - The circles are external to each other: c > a+b.
[a2 - b2 + c2]2 - 4a2c2 = (c+a+b)(c+a-b)(c-a+b)(c-a-b) > 0
The distance between P and Q is √([a2 - b2 + c2]2 - 4a2c2)/c
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