smc

Let $OM=x$, $ON=y$. By the Pythagorean Theorem on $△XON,MOY$ respectively, $$\begin{array}{rl}(2x{)}^2+y^2& =19^2\\ x^2+(2y{)}^2& =22^2\end{array}$$ Summing these gives $5x^2+5y^2=845⟹x^2+y^2=169$. By the Pythagorean Theorem again, we have $$(2x{)}^2+(2y{)}^2=X{Y}^2⟹XY=\sqrt{4(x^2+y^2)}=\sqrt{4(169)}=\sqrt{676}=\boxed{(\mathrm{B})26}$$ Alternatively, we could note that since we found $x^2+y^2=169$, segment $MN=13$. Right triangles $△MON$ and $△XOY$ are similar by Leg-Leg with a ratio of $\frac{1}{2}$, so $XY=2(MN)=\boxed{26}$