Problem: Given circles $\omega_1$ and $\omega_2$ intersecting at points $X$ and $Y$,
let $\ell_1$ be a line through the center of $\omega_1$
intersecting $\omega_2$ at points $P$ and $Q$
and let $\ell_2$ be a line through the center of $\omega_2$
intersecting $\omega_1$ at points $R$ and $S$.
Prove that if $P$, $Q$, $R$, and $S$ lie on a circle
then the center of this circle lies on line $XY$.

Solution: Let $r_1$, $r_2$, $r_3$ denote the
circumradii of $\omega_1$, $\omega_2$, and $\omega_3$, respectively,
where $\omega_3$ is the circle through $P$, $Q$, $R$, $S$.

\begin{center}
\begin{asy}
pair O_1 = dir(220);
pair O_2 = dir(320);
pair O = dir(110);
pair T = foot(O, O_1, O_2);
pair X = midpoint(O--T);
pair Y = 2*T-X;

filldraw(CP(O_1, X), opacity(0.1)+lightcyan, lightblue);
filldraw(CP(O_2, X), opacity(0.1)+lightcyan, lightblue);

draw(O--Y, red+dashed);

pair K = foot(O_1, O, O_2);
pair P = IP(O_1--K, CP(O_2, X));
pair Q = 2*K-P;
draw(O_1--Q, red);
draw(O--O_2, heavycyan);

pair L = foot(O_2, O, O_1);
pair R = IP(O_2--L, CP(O_1, X));
pair S = 2*L-R;
draw(O_2--S, red);
draw(O--O_1, heavycyan);

draw(arc(O, abs(P-O), 180, 360), heavygreen);

dot("$O_1$", O_1, dir(O_1));
dot("$O_2$", O_2, dir(O_2));
dot("$O$", O, dir(O));
dot("$X$", X, dir(100));
dot("$Y$", Y, dir(Y));
dot("$P$", P, dir(120));
dot("$Q$", Q, dir(Q));
dot("$R$", R, dir(40));
dot("$S$", S, dir(S));

/* TSQ Source:

O_1 = dir 220
O_2 = dir 320
O = dir 110
T := foot O O_1 O_2
X = midpoint O--T R100
Y = 2*T-X

CP O_1 X 0.1 lightcyan / lightblue
CP O_2 X 0.1 lightcyan / lightblue

O--Y red dashed

K := foot O_1 O O_2
P = IP O_1--K CP O_2 X R120
Q = 2*K-P
O_1--Q red
O--O_2 heavycyan

L := foot O_2 O O_1
R = IP O_2--L CP O_1 X R40
S = 2*L-R
O_2--S red
O--O_1 heavycyan

!draw(arc(O, abs(P-O), 180, 360), heavygreen);

*/
\end{asy}
\end{center}

We wish to show that $O_3$ lies on the radical axis of $\omega_1$ and $\omega_2$.
Let us encode the conditions using power of a point.
Because $O_1$ is on the radical axis of $\omega_2$ and $\omega_3$,
\begin{align*}
  \opname{Pow}_{\omega_2}(O_1) &= \opname{Pow}_{\omega_3}(O_1) \\
  \implies O_1O_2^2 - r_2^2 &= O_1O_3^2 - r_3^2.
  \intertext{Similarly, because $O_2$ is on the radical axis of $\omega_1$ and $\omega_3$, we have}
  \opname{Pow}_{\omega_1}(O_2) &= \opname{Pow}_{\omega_3}(O_2) \\
  \implies O_1O_2^2 - r_1^2 &= O_2O_3^2 - r_3^2. \\
  \intertext{Subtracting the two gives}
  (O_1O_2^2-r_2^2)-(O_1O_2^2-r_1^2) &= (O_1O_3^2-r_3^2)-(O_2O_3^2-r_3^2) \\
  \implies r_1^2-r_2^2 &= O_1O_3^2-O_2O_3^2 \\
  \implies O_2O_3^2 - r_2^2 &= O_1O_3^2 - r_1^2 \\
  \implies \opname{Pow}_{\omega_2}(O_3) &= \opname{Pow}_{\omega_1} (O_3)
\end{align*}
as desired.