Given two circles C and C' we say that C bisects C' if their common chord is a diameter of C'. Show that for any two circles which are not concentric, there are infinitely many circles which bisect them both. Find the locus of the centers of the bisecting circles.
Solution
Let C, C' have center O, O' respectively and radius r, r' respectively. Let a circle center P bisect C. Suppose it meets C at A and B. Then AB is perpendicular to OP and is a diameter of C. Hence PA2 = OP2 + r2. Conversely, the circle center P, radius √(OP2 + r2) bisects C. So P will bisect C and C' iff OP2 + r2 = OP'2 + r'2.
It is well-known that the locus of points P' with equal tangents to C and C' is the radical axis. Call the radical axis R. For a point P' on the radical axis we have P'O2 - r2 = P'O'2 - r'2. If we reflect P' in the perpendicular bisector of OO' to get P, then PO = P'O' and PO' = P'O, so PO'2 - r2 = PO2 - r'2 and hence PO2 + r2. Call the reflection of the R in the perpendicular bisector of OO' the line R'. We have established that points on R' form part of the locus. Conversely, if P' is such that there is a circle center P' bisecting both circles, then OP'2 + r2 = O'P'2 + r'2, so if P is the reflection of P' then OP2 - r2 = OP'2 - r'2 and hence P lies on the radical axis R. Hence P' must lie on R'.
Radical axis
We have PT2 = PO2 - r2 = PX2 + OX2 - r2 , and similarly PT'2 = PX2 + O'X2 - r'2. So PT = PT' iff OX2 - r2 = O'X2 - r'2. There is evidently a unique point X for which that is true, so the locus of such P is the line through X perpendicular to OO'
If the circles intersect, then the point X evidently lies on the line joining the two common points, because OX2 - r2 = -XY2 = O'X2 - r'2. In any case the midpoint of each common tangent evidently lies on the line, so that provides a way of constructing it.
© John Scholes
jscholes@kalva.demon.co.uk
24 Oct 2002