Four distinct points A_{1}, A_{2}, B_{1}, B_{2} have the property that any circle through A_{1} and A_{2} has at least one point in common with any circle through B_{1} and B_{2}. Show that the four points are collinear or lie on a circle.

**Solution**

The trick is to take *concentric* circles. If A_{1}A_{2} is not parallel to B_{1}B_{2} then their perpendicular bisectors must intersect at some point O. Take circles centre O through A_{1}, A_{2} and through B_{1}, B_{2}. These must either coincide, in which case the 4 points lie on a circle, or have no points in common.

In the case where A_{1}A_{2} is parallel to B_{1}B_{2}. Assume they do not coincide (otherwise the 4 points would be collinear). Then take a point O on the perpendicular bisector of A_{1}A_{2} on the opposite side of A_{1}A_{2} to B_{1}B_{2} and sufficiently distant from A_{1}A_{2} that the circle through A_{1}, A_{2} centre O only extends less than halfway towards the line B_{1}B_{2}. Similarly, take a circle through B_{1}, B_{2} which extends less than halfway towards A_{1}A_{2}. Then these circles will not meet.

© John Scholes

jscholes@kalva.demon.co.uk

25 Jan 2002