Problem: Circle $\omega$, centered at $X$, is internally tangent to circle $\Omega$,
centered at $Y$, at $T$.
Let $P$ and $S$ be variable points on $\Omega$ and $\omega$,
respectively, such that line $PS$ is tangent to $\omega$ (at $S$).
Determine the locus of $O$ --- the circumcenter of triangle $PST$.

Solution: The answer is a circle centered at $Y$
with radius $\sqrt{YX \cdot YT}$,
minus the two points on line $XY$ itself.

We let $PS$ meet $\Omega$ again at $P'$,
and let $O'$ be the circumcenter of $\triangle TPS'$.
Note that $O'$, $X$, $O$ are collinear on the perpendicular
bisector of line $\ol{TS}$
Finally, we let $M$ denote the arc midpoint of $PP'$
which lies on line $TS$ (by homothety).

\begin{center}
\begin{asy}
import graph; size(9cm);
pen qqffff = rgb(0.,1.,1.); pen qqwuqq = rgb(0.,0.39215,0.); pen ffdxqq = rgb(1.,0.84313,0.); pen yqqqqq = rgb(0.50196,0.,0.);
pair X = (-2.,0.), Y = (2.,0.), T = (-9.50185,0.), P = (7.49229,10.10580), O = (5.46686,-5.82994), M = (-0.78960,11.15843);
pair S = (-3.81946,7.27786);
pair Pp = (-7.60181,6.33228);
pair Op = (-4.49671,1.94937);
fill(T--S--P--cycle, palegreen);
fill(T--Op--Y--cycle, palegreen);
fill(T--S--Pp--cycle, palecyan);
fill(T--O--Y--cycle, palecyan);
draw(circle(X, 7.50185), linewidth(0.6) + red);
draw(circle(Y, 11.50185), linewidth(0.6) + red);
draw(circle(Y, 6.78287), linewidth(0.6) + qqwuqq);
draw(T--Op--Y--O--T--Y, gray);
draw(T--Pp--M--P--T--M, black);
draw(Pp--P, black);
draw(Op--O, gray);
dot("$X$", X, dir((8.005, 21.422)));
dot("$Y$", Y, dir((8.687, 21.422)));
dot("$T$", T, dir(T-Y));
dot("$S$", S, dir(135));
dot("$P$", P, dir(P-Y));
dot("$O$", O, dir(-45));
dot("$P'$", Pp, dir(Pp-Y));
dot("$O'$", Op, dir(40));
dot("$M$", M, dir((8.912, 22.013)));
\end{asy}
\end{center}
By three applications of Salmon theorem, we have the following spiral
similarities all centered at $T$:
\begin{align*}
  \triangle TSP &\overset{+}{\sim} \triangle TO'Y \\
  \triangle TP'S &\overset{+}{\sim} \triangle TYO \\
  \triangle TP'P &\overset{+}{\sim} \triangle TO'O.
\end{align*}
However, the shooting lemma also gives us two similarities:
\begin{align*}
  \triangle TP'M &\overset{+}{\sim} \triangle TSP \\
  \triangle TMP &\overset{+}{\sim} \triangle TP'S.
\end{align*}
Putting everything together, we find that
\[ TP'MP \overset{+}{\sim} TO'YO. \]
Then by shooting lemma, $YO'^2 = YX \cdot YT$, so $O$
indeed lies on the claimed circle.

As the line $\ol{O'O}$ may be any line through $X$ other than line $XY$
(one takes $S$ to be the reflection of $T$ across this line)
one concludes the only two non-achievable points are
the diametrically opposite ones on line $XY$ of this circle
(because this leads to the only degenerate situation where $S=T$).