Let $P$ and $Q$ be on arc $\widehat{AC}$ of $(ABC)$ such that $\angle CAP = \angle ACQ= \angle BAX = \angle BCY$. Now stick everything on the complex plane, with $a,b,c$ on the unit circle as usual. We can parameterize $p=az, q = \frac{c}{z}$ for some $z$ on the unit circle. Now by spiral similarity we have $$\frac{x}{y} = \frac{a + (za-a)\cdot \frac{b-a}{c-a}}{c + (\frac{c}{z}-c)\cdot \frac{b-c}{a-c}} = -\frac{zab-za^2-ab+ac}{\frac{bc}{z}-\frac{c^2}{z}-bc+ac} = -\frac{a}{c}\cdot \frac{zb-za-b+c}{\frac{b}{z}-\frac{c}{z}-b+a} = -\frac{a}{c}\cdot \frac{z(b-a) - (b-c)}{\frac{1}{z}(b-c) - (b-a)} = -\frac{a}{c}\cdot -z$$ which has magnitude $1$, as desired.

Let M be the second intesction of BX and （ABC），N be the second intesction of BY and （ABC）， we have $\angle AXM =\angle B =\angle CYN, \angle AMX =\angle C, \angle CNY =\angle A$ so$\triangle AXM \sim \triangle ABC \sim \triangle NYC$ but AM=CN, so we have $\triangle AXM \cong \triangle NYC$ $\therefore AX=NY, XM=CY$ it's trivial that $\triangle AXB \sim \triangle CYB$, so $\dfrac {AX}{BX}=\dfrac {CY}{BY}$ $\therefore BX \cdot XM =CY \cdot NY$, whicht means $R^2-OX^2=R^2-OY^2$,so $OX=OY$ Q.E.D