A circle $\omega$ touched lines $a$ and $b$ at points $A$ and $B$ respectively. An arbitrary tangent to the circle meets $a$ and $b$ at $X$ and $Y$ respectively. Points $X'$ and $Y'$ are the reflections of $X$ and $Y$ about $A$ and $B$ respectively. Find the locus of projections of the center of the circle to the lines $X'Y'$.
We claim that the locus is the reflection of $w$ about $AB$. Let $XY$ touch $w$ at $T$ and let $O'$ be the reflection of $O$, the center of the circle, about $AB$. Place $w$ as the unit circle. The lowercase letter represents the complex number. Then $x=\frac{2at}{a+t}$ and so $x'=\frac{2a^2}{a+t}$. So the foot of $O$ onto $X'Y'$ is then given by $$\frac{\overline{x}'y'-x'\overline{y}'}{2(\overline{x}'-\overline{y}')}=\frac{4\overline{a}tb^2-4\overline{b}ta^2}{4\overline{a}t(b+t)-4\overline{b}t(a+t)}=\frac{a^2+ab+b^2}{t+a+b}$$Taking the displacement of this projection with $O'$ gives $$\frac{ab+at+bt}{a+b+t}$$Also we have that $$\frac{ab+at+bt}{a+b+t}\cdot \frac{\overline{a}\overline{b}+\overline{a}\overline{t}+\overline{b}\overline{t}}{\overline{t}+\overline{a}+\overline{b}}=\frac{ab+at+bt}{a+b+t}\cdot \frac{a+b+t}{ab+at+bt}=1$$Furthermore, every point on the locus is achievable as if $|z|=1$ then $$z=\frac{ab+at+bt}{a+b+t}\iff t=\frac{-ab+az+bz}{a+b-z}$$But this fraction itself has magnitude $1$, as desired.