Answer: 717

Small Hint:

The tangent-chord angle gives $\angle ABT = 180^\circ - C,$ so $TX = 16 \sin C;$ and $AXTY$ is cyclic with diameter $AT,$ so $XY = AT \sin A$

Big Hint:

Place $B$ and $C$ at $(\mp 11, 0),$ so $T = (0, -\sqrt{135}).$ Then $TX^2 + TY^2$ and $XY^2$ are both linear in the $y$-coordinate of $A.$

Solution:

By the tangent-chord angle, $\angle TBC = A,$ so $\angle ABT = B + A = 180^\circ - C$ and $TX = TB \sin\angle ABT = 16 \sin C;$ similarly $TY = 16 \sin B.$ Also $\angle AXT = \angle AYT = 90^\circ,$ so $A,$ $X,$ $T,$ and $Y$ lie on a circle with diameter $AT,$ whence $XY = AT \sin A.$ Using the law of sines, $\sin A = \frac{11}{R},$ $\sin B = \frac{AC}{2R},$ and $\sin C = \frac{AB}{2R},$ so the given condition becomes $$\begin{aligned} &\frac{64\left(AB^2 + AC^2\right) + 121\,AT^2}{R^2} \\ &= 1143. \end{aligned}$$

Place $B = (-11, 0)$ and $C = (11, 0).$ Since $TB = 16$ and $T$ lies on the perpendicular bisector of $BC,$ we get $T = (0, -\sqrt{135}).$ The circumcenter is $O = (0, k)$ with $OB \perp BT,$ which gives $121 - k\sqrt{135} = 0,$ so $k = \frac{121}{\sqrt{135}}$ and $R^2 = 121 + k^2 = \frac{30976}{135}.$ For $A = (x, y)$ on $\omega,$ expanding $x^2 + (y - k)^2 = R^2$ gives $x^2 + y^2 = \frac{242}{\sqrt{135}}\,y + 121.$ Therefore $$\begin{aligned} AB^2 + AC^2 &= 2(x^2 + y^2) + 242 \\ &= \frac{484}{\sqrt{135}}\,y + 484, \end{aligned}$$ $$\begin{aligned} AT^2 &= x^2 + y^2 \\ &\quad {}+ 2\sqrt{135}\,y + 135 \\ &= \frac{512}{\sqrt{135}}\,y + 256. \end{aligned}$$

Substituting, $64(AB^2 + AC^2) + 121\,AT^2$ $= \frac{92928}{\sqrt{135}}\,y + 61952$ $= 1143 \cdot \frac{30976}{135}$ yields $y = \frac{291}{\sqrt{135}}.$ Then $AT^2 = \frac{512 \cdot 291}{135} + 256 = \frac{183552}{135},$ and $$\begin{aligned} XY^2 &= AT^2 \sin^2 A \\ &= \frac{121\,AT^2}{R^2} \\ &= \frac{121 \cdot 183552}{30976} \\ &= \frac{183552}{256} = 717. \end{aligned}$$