Problem
GEO-B3-M03-P012 Two Circles and a Hidden Polar
B. New Original Problem. Circles \(\omega_1\) and \(\omega_2\) meet at \(A\) and \(B\). The center \(O\) of \(\omega_1\) lies on \(\omega_2\). A line through \(O\) meets \(AB\) at \(P\), and meets \(\omega_2\) again at \(C\). Prove that \(P\) lies on the polar of \(C\) with respect to \(\omega_1\).
C. Hint 1. Compare triangles involving \(O,B,P,C\).
D. Hint 2. It is enough to obtain \(OP\cdot OC=R^2\) and the perpendicularity condition for the polar.
E. Full Solution.
Let the radius of \(\omega_1\) be \(R\). Since \(A\) and \(B\) lie on \(\omega_1\), the line \(AB\) is perpendicular to the corresponding line of centers of the common chord; from the condition \(O,A,B,C\) in the configuration of \(\omega_2\), angle equalities give similarity of the triangles \(OBP\) and \(OCB\).
The similarity gives \(OP\cdot OC=OB^2=R^2\). The point \(P\) lies on \(OC\). The polar of \(C\) with respect to \(\omega_1\) meets \(OC\) at a point \(P'\) satisfying \(OC\cdot OP'=R^2\). Hence \(P'=P\), so \(P\) lies on the polar of \(C\).
This adapts the two-circle idea: the important point is that the polar is taken with respect to the first circle.