Let $\omega_1$ and $\omega_2$ be the circles, with centers $O_1$ and $O_2$. Let $O$ be the midpoint of $\overline{O_1O_2}$. Call a line pink if it cuts equal chords in $\omega_1$ and $\omega_2$. Key Claim. Let $\ell$ be a pink line. Then the projection of $O$ onto $\ell$ lies on the radical axis of $\omega_1$ and $\omega_2$. Proof. Let $\ell$ meet $\omega_1$ and $\omega_2$ at $\{A, B\}$ and $\{C, D\}$, and suppose that $\overrightarrow{AB} = \overrightarrow{CD}$. Let $P_1$, $P_2$, $P$ be the projections of $O_1$, $O_2$, $O$ onto $\ell$. Since $OO_1 = OO_2$, we get $PP_1 = PP_2$. In addition $P_1A = P_1B = P_2C = P_2D$, so we have $PA \cdot PB = PC \cdot PD$. Thus $P$ lies on the radical axis. $\square$ Now, consider three pink lines: the projections of $O$ onto these lines are collinear, so $O$ lies on the circumcircle of the triangle determined by these lines.