Solution:

Reflecting $H$ over line $BC$ lands on $\omega,$ so the circumcircle of $HBC$ is the reflection of $\omega$ over $BC.$ Take the circumcenter $O$ as the origin, so that $H = A + B + C$ as vectors. If $M$ is the midpoint of $\overline{BC},$ then $OM \perp BC,$ so the reflected center is $2M - O = B + C = H - A.$ Tangency at $H$ means $XY$ is perpendicular to the radius from $B + C$ to $H,$ which is the vector $A:$ the chord $XY$ is perpendicular to $OA.$

Place $A = (0, R)$ so that $XY$ is horizontal at height $h,$ with $H = (x_0, h).$ The half-chord length is $\sqrt{R^2 - h^2},$ and $HX = 2,$ $HY = 6$ give $\sqrt{R^2 - h^2} = 4$ with $|x_0| = 2.$ From $HA = 3:$ $4 + (R - h)^2 = 9,$ so $R - h = \sqrt{5}.$ Then $$16 = R^2 - h^2 = (R - h)(R + h) = \sqrt{5}\left(2R - \sqrt{5}\right),$$ giving $R = \frac{21}{2\sqrt{5}}.$

Now $B + C = H - A = (\pm 2, -\sqrt{5}),$ so $M = \left(\pm 1, -\frac{\sqrt{5}}{2}\right)$ and $OM = \frac{3}{2},$ whence $BC = 2\sqrt{R^2 - \frac{9}{4}} = 2\sqrt{\frac{99}{5}}.$ The distance from $A$ to line $BC$ (through $M,$ perpendicular to $OM$) is $\frac{|A \cdot M - OM^2|}{OM} = \frac{21/4 + 9/4}{3/2} = 5,$ using $A \cdot M = -\frac{\sqrt{5}R}{2} = -\frac{21}{4}.$ Hence $$[ABC] = \frac{1}{2} \cdot 2\sqrt{\frac{99}{5}} \cdot 5 = \sqrt{495} = 3\sqrt{55},$$ and $m + n = 3 + 55 = 58.$