If the common tangents of $x^2+y^2=r^2$ and $\frac{x^2}{16}+\frac{y^2}{9}=1$ form a square, then the area (in sq. units) of the square is

Let, the equation of tangent in slope form for the circle be $y=mx\pm r\sqrt{1+m^2}$

and the equation of tangent in slope form for the ellipse be $y=mx\pm \sqrt{16m^2+9}$

Now, for common tangets,
$y=mx\pm r\sqrt{1+m^2}$ is same as $y=mx\pm \sqrt{16m^2+9}$
$\Rightarrow r^2+r^2{m}^2=16m^2+9\Rightarrow \left(r^2-16\right)m^2+\left(r^2-9\right)=0$

Since, here $m_1{m}_2=-1$ 
$\Rightarrow 2r^2=25\Rightarrow r=\frac{5}{\sqrt{2}}$

So, the side of the square $=2r=5\sqrt{2}$
$\Rightarrow$ area of the square is $50$ sq. units