Problem
GEO-B3-M01-P003 A Line Becomes a Circle
Line \(l\) does not pass through point \(O\). The perpendicular \(OC\) is dropped to \(l\), and \(C^*\) is the image of \(C\) under inversion centered at \(O\). Prove that the image of line \(l\) is the circle with diameter \(OC^*\).
Hint 1. Take an arbitrary point \(M\) on \(l\).
Hint 2. Prove that \(\angle OM^*C^*=90^\circ\).
E. Full solution. Let \(M\in l\). Then \(\angle OCM=90^\circ\). By the similarity lemma for inverse triangles, \(\triangle OCM\sim\triangle OM^*C^*\). Hence \(\angle OM^*C^*=90^\circ\). Therefore all points \(M^*\) lie on the circle with diameter \(OC^*\). Conversely, if \(X\) lies on this circle, line \(OX\) meets \(l\) at a point \(M\), and the same similarity gives \(X=M^*\).
A key training problem on the image of a line; without it, later inversion arguments are impossible.