In the plane points $A,B,C$ are marked in blue and points $P,Q$ are marked in red (no 3 marked points lie on a line, and no 4 marked points lie on a circle). A circle is called separating if all points of one color are inside it, and all points of the other color are outside of it. Denote by $O$ the circumcenter of $ABC$ and by $R$ the circumradius of $ABC$. Prove that exactly one of the following holds: There exists a separating circle; There exists a point $X$ on the segment $PQ$ which also lies inside the triangle $ABC$, for which $PX\cdot XQ = R^2-OX^2$.