Solution:

Consider a rotation about $X$ by $90^{\circ}$ followed by a homothety with ratio $\frac{1}{3}$ that sends $B$ to $A$. This sends $\omega$ to $\omega'$ with radius $\frac{1}{3}$ of the radius of $\omega$ and center $O'$. Since $A$ is the image of $B$ under this rotation, we know $A$ lies on both circles; the same argument shows $Y$ must lie on both circles. Thus, $Y$ is the reflection of $A$ over $OO'$. In particular, this means that $XY = AX'$, where $X'$ is the reflection of $X$ over $OO'$.

Let $M$ be the midpoint of $AB$. Note that because $\triangle OXO'$ $\sim$ $\triangle BXA$, we also have $\triangle XOX'$ $\sim$ $\triangle XMA$, as they are both isosceles and $\angle XOX' = 2\angle XOO' = 2\angle XBA = \angle XMA$. This implies that $\triangle XOM \sim \triangle XX'A$. Thus, we know that $AX' = OM \cdot \frac{XA}{XM} = \frac{2}{\sqrt{10}} OM$. It remains to compute $OM$; noting that the distance between $O$ and the foot from $X$ to $AB$ is $\frac{2}{5}\sqrt{10}$, and that the altitude of $\triangle AXY$ has length $\frac{3}{10}\sqrt{10}$, we get that the distance from $O$ to $AB$ is

$$\frac{3}{10}\sqrt{10} + \sqrt{\left(2\sqrt{2}\right)^2 - \left(\frac{2}{5}\sqrt{10}\right)^2} = \frac{11}{10}\sqrt{10}$$

by the Pythagorean theorem, which means that $XY = AX' = \left\lfloor \frac{11}{5} \right\rfloor$.

Solution:

Let $M$ be the midpoint of $AB$. We will find $MO$ first.

Let the internal bisector of $\angle A X B$ intersect $\odot (A X B)$ at $P$. From Ptolemy, $X P = 2\sqrt{2} = X O$. Let $X'$ be the foot of altitude from $X$ to $M O$. Observe that $O$ is the reflection of $P$ across $X X'$. By the area of $\triangle A X B$, we have $X'M = \frac{3}{\sqrt{10}}$. Therefore,

$$M O = O X' + X'M = X'P + X'M = 2X'M + M P = \frac{11\sqrt{10}}{10}.$$

By the spiral similarity $\triangle X A B \mapsto \triangle X Y Z$, we have that $A Y \perp B Z$ and $\angle A O B + \angle Y O Z = 90^{\circ}$ Therefore, $\triangle X A M \sim \triangle X Y N$ where $N$ is the midpoint of $Y Z$. Thus, $Y Z = \frac{11\sqrt{10}}{5}$ and $X Y = \left[\frac{11}{5}\right]$