jmc

For the hyperbola $\mathcal{H},$ we have $a=3$ and $c=5,$ so $b=\sqrt{c^2-a^2}=4.$ Thus, the hyperbola has equation $$\frac{x^2}{3^2}-\frac{y^2}{4^2}=1,$$or $$16x^2-9y^2=144.$$Meanwhile, the equation for the circle is $x^2+y^2=16.$ To find the points of intersection, we solve these two equations simultaneously. Adding $9$ times the second equation to the first equation gives $25x^2=288,$ so $x=\pm \frac{12\sqrt{2}}{5}.$ Then we have $$y^2=16-x^2=16-\frac{288}{25}=\frac{112}{25},$$so $y=\pm \frac{4\sqrt{7}}{5}.$ Therefore, the four points of intersection form a rectangle with side lengths $\frac{24\sqrt{2}}{5}$ and $\frac{8\sqrt{7}}{5},$ so its area is $\frac{24\sqrt{2}}{5}\cdot \frac{8\sqrt{7}}{5}=\boxed{\frac{192\sqrt{14}}{25}}.$