Given that 

$\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi }{4}$

Let $z = x + iy$
$\Rightarrow \arg \left(\frac{x+iy-2}{x+iy+2}\right)=\frac{\pi }{4}$

$\Rightarrow \arg \left(\left(\frac{x+iy-2}{x+iy+2}\right)\left(\frac{x+2-iy}{x+2-iy}\right)\right)=\frac{\pi }{4}$

$\Rightarrow \arg \left(\frac{x^2+y^2-4+4iy}{(x+2{)}^2+y^2}\right)=\frac{\pi }{4}$

$\Rightarrow \frac{x^2+y^2-4+4iy}{(x+2{)}^2+y^2}=\tan \left(\frac{\pi }{4}\right)$

$\Rightarrow \frac{x^2+y^2-4}{4y}=1$

$x^2+(y-2{)}^2=8$    ……….(1)

Equation $\left(1\right)$  represents circle, whose centre $(0,2)$, radius $2\sqrt{2}$.

Minimum distance $d=|A C-r|$
Given point $A(9\sqrt{2},2)$

$=|AC-r|=\mid (9\sqrt{2}-2\sqrt{2})=7\sqrt{2}$

$\Rightarrow$ Square of distance $=d^2=(7\sqrt{2}{)}^2=98$