[asy][asy] size(6cm); defaultpen(fontsize(9pt)); import geometry; pair A, B, C, D, O, P, X; guide w1, w2, c; w1 = unitcircle; w2 = circle((2, 0), 1.6); pair[] op = intersectionpoints(w1, w2); O = op[1]; P = op[0]; c = circle(O, 0.6); pair[] ab = intersectionpoints(c, w1), cd = intersectionpoints(c, w2); A = ab[1]; B = ab[0]; C = cd[0]; D = cd[1]; X = extension(A, C, B, D); draw(w1^^w2, blue); draw(c, red); draw(A--X--D); draw(line(P, X), heavygreen+dashed); dot("$A$", A, dir(-120)); dot("$B$", B, dir(25)); dot("$C$", C, dir(150)); dot("$D$", D, dir(-95)); dot("$P$", P, dir(93)); dot("$O$", O, dir(-105)); dot("$X$", X, dir(80)); [/asy][/asy] Let $\omega_1$ and $\omega_2$ intersect again at $P$. We claim that $\angle OPX = 90^\circ$. Let $Q = AB \cap CD$. Evidently $P$ and $Q$ are inverses wrt the variable circle; in particular, $O-P-X$ are collinear and $P$ lies on the polar of $Q$. By Brocard's theorem, so does $X$. Therefore, $\overline{PX}$ is the polar of $Q$, as desired.