Problem
GEO-B3-M01-P004 A Circle Through the Center
Circle \(\omega\) passes through the inversion center \(O\). Line \(OO_1\), where \(O_1\) is the center of \(\omega\), meets \(\omega\) again at \(A\). Prove that the image of \(\omega\) is the line perpendicular to \(OA\) through \(A^*\).
Hint 1. For \(M\in\omega\), angle \(OMA\) is right.
Hint 2. Inversion is its own inverse: use the result about the image of a line.
E. Full solution. Since \(OA\) is a diameter of \(\omega\), for every \(M\in\omega\) we have \(\angle OMA=90^\circ\). Under inversion, points \(M\) map to points \(M^*\) on rays \(OM\). By the similarity of inverse triangles, the condition \(\angle OMA=90^\circ\) means that \(M^*\) lies on the line through \(A^*\) perpendicular to \(OA\). The reverse inclusion follows from the invertibility of inversion.
This is the inverse of the previous problem; it is important to see both directions of the transformation.