ABCD is a fixed cyclic quadrilateral with AB not parallel to CD. Find the locus of points P for which we can find circles through AB and CD touching at P.
Solution
Answer: Let the lines AB and CD meet at X. Let R be the length of a tangent from X to the circle ABCD. The locus is the circle center X radius R. [Strictly you must exclude four points unless you allow the degenerate straight line circles.]
Let X be the intersection of the lines AB and CD. Let R be the length of a tangent from X to the circle ABCD. Let C0 be the circle center X radius R. Take any point P on C0. Then considering the original circle ABCD, we have that R2 = XA·XB = XC·XD, and hence XP2 = XA·XB = XC·XD.
If C1 is the circle through C, D and P, then XC.XD = XP2, so XP is tangent to the circle C1. Similarly, the circle C2 through A, B and P is tangent to XP. Hence C1 and C2 are tangent to each other at P. Note that if P is one of the 4 points on AB or CD and C0, then this construction does not work unless we allow the degenerate straight line circles AB and CD.
So we have established that all (or all but 4) points of C0 lie on the locus. But for any given circle through C, D, there are only two circles through A, B which touch it (this is clear if you consider how the circle through A, B changes as its center moves along the perpendicular bisector of AB), so there are at most 2 points on the locus lying on a given circle through C, D. But these are just the two points of intersection of the circle with C0. So there are no points on the locus not on C0.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002