jmc

Let $O$ be the center of the circle, and $r$ its radius, and let $X^{\prime }$ and $Y^{\prime }$ be the points diametrically opposite $X$ and $Y$, respectively. We have $O{X}^{\prime }=O{Y}^{\prime }=r$, and $\angle X^{\prime }O{Y}^{\prime }=90^{\circ }$. Since triangles $X^{\prime }O{Y}^{\prime }$ and $BAC$ are similar, we see that $AB=AC$. Let $X^{\prime \prime }$ be the foot of the altitude from $Y^{\prime }$ to $\overline{AB}$. Since $X^{\prime \prime }B{Y}^{\prime }$ is similar to $ABC$, and $X^{\prime \prime }{Y}^{\prime }=r$, we have $X^{\prime \prime }B=r$. It follows that $AB=3r$, so $r=2$.



Then, the desired area is the area of the quarter circle minus that of the triangle $X^{\prime }O{Y}^{\prime }$. And the answer is $\frac{1}{4}\pi r^2-\frac{1}{2}{r}^2=\boxed{\pi -2}$.