Problem
GEO-B3-M01-P006 Orthogonal Circle
Circle \(\gamma\) is orthogonal to the circle of inversion centered at \(O\) with radius \(R\). Prove that \(\gamma\) maps to itself.
Hint 1. Compute the power of \(O\) with respect to \(\gamma\).
Hint 2. If line \(OX\) meets \(\gamma\) at \(X\) and \(Y\), what is \(OX\cdot OY\)?
E. Full solution. Let \(C\) and \(r\) be the center and radius of \(\gamma\). Orthogonality gives \(OC^2=R^2+r^2\). Hence the power of \(O\) with respect to \(\gamma\) equals \(OC^2-r^2=R^2\). If a line through \(O\) meets \(\gamma\) at \(X\) and \(Y\), then \(OX\cdot OY=R^2\). Therefore \(Y\) is the image of \(X\) under the inversion. Thus every point of \(\gamma\) maps to a point of the same circle.
Orthogonal circles are one way to make an important circle fixed.