Problem
GEO-B3-M01-P014 A Circle Equally Inclined to Two Circles
Circles \(\omega_1\) and \(\omega_2\) meet at points \(A\) and \(B\). Construct a circle through \(A\) that cuts \(\omega_1\) and \(\omega_2\) at equal angles. Justify why there are usually two such circles.
Hint 1. Invert centered at \(A\).
Hint 2. Circles \(\omega_1\) and \(\omega_2\) become two lines through \(B^*\).
E. Full solution. Perform an arbitrary inversion centered at \(A\). Circles \(\omega_1\) and \(\omega_2\), passing through the inversion center, become two lines \(l_1\) and \(l_2\) meeting at \(B^*\). The desired circle also passes through \(A\), so its image is a line \(m\). Inversion preserves intersection angles, hence the equal-angle condition becomes: line \(m\) makes equal angles with \(l_1\) and \(l_2\). Therefore \(m\) must be one of the two angle bisectors of \(l_1\) and \(l_2\). Inverting these two bisectors back gives the two desired circles through \(A\).
This problem is useful because after inversion a complicated circle becomes a simple angle bisector between two lines.