USAMO, 2009

Let $\omega$ denote the circumcircle of $P, Q, R, S$ and let $O$ denote the center of $\omega$. Line $XY$ is the radical axis of circles $\omega_1$ and $\omega_2$. It suffices to show that $O$ has equal power to the two circles; that is, to show that
$$OO_1^2 - O_1S^2 = OO_2^2 - O_2Q^2 \quad \text{or} \quad OO_1^2 + O_2Q^2 = OO_2^2 + O_1S^2.$$
Let $M$ and $N$ be the intersections of lines $O_2O, \ell_1$ and $O_1O, \ell_2$. Because circles $\omega$ and $\omega_2$ intersect at points $P$ and $Q$, we have $PQ \perp OO_2$ (or $\ell_1 \perp OO_2$). Hence
$$OO_1^2 - OQ^2 = (OM^2 + MO_1^2) - (OM^2 + MQ^2) = (O_2M^2 + MO_1^2) - (O_2M^2 + MQ^2) = O_2O_1^2 - O_2Q^2$$
or
$$O_2O_1^2 + OQ^2 = OO_1^2 + O_2Q^2.$$
Likewise, we have $O_2O_1^2 + OS^2 = OO_2^2 + O_1S^2$. Because $OS = OQ$, we obtain that $OO_1^2 + O_2Q^2 = OO_2^2 + O_1S^2$, which is what was to be proved.

We maintain the notations of the first solution. Three pairs of circles $(\omega, \omega_1)$, $(\omega_1, \omega_2)$, $(\omega_2, \omega)$ meet at three pairs of points $(R, S)$, $(X, Y)$, $(P, Q)$, respectively; that is, lines $RS, XY, PQ$ are the respective radical axes of these pairs of circles. Thus, these three radical axes must be concurrent at the radical center, denoted by $H$, of these three circles. In particular, it follows that $H, X, Y$ lie on a line, denoted by $\ell$, and $\ell \perp O_1O_2$.

On the other hand, $O_1M \perp O_2O$ and $O_2N \perp O_1O$. Hence $H$ is the orthocenter of triangle $OO_1O_2$, from which it follows that $OH \perp O_1O_2$. Therefore, $O$ lies on $\ell$; that is, $X, P, Q$ are collinear.