Problem
GEO-B3-M01-P005 A Circle Not Through the Center
Circle \(\omega\) does not pass through the inversion center \(O\). The line joining \(O\) to the center of \(\omega\) meets \(\omega\) at \(A\) and \(B\). Prove that the image of \(\omega\) is the circle with diameter \(A^*B^*\).
Hint 1. For any \(M\in\omega\), angle \(AMB\) is right.
Hint 2. Transfer this right angle to the images.
E. Full solution. Points \(A\) and \(B\) are endpoints of a diameter of \(\omega\), so \(\angle AMB=90^\circ\) for every \(M\in\omega\). By the similarity of inverse triangles, the angle between \(A^*M^*\) and \(B^*M^*\) is also right. Hence \(M^*\) lies on the circle with diameter \(A^*B^*\). Since inversion is reversible, the whole circle is the image of \(\omega\).
This is the first problem where one must choose the correct diameter of the original circle.