Question

A point $z$ moves in the complex plane such that $\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi }{4}$, then the minimum value of $|z-9\sqrt{2}-2i{|}^2$ is equal to ____

Answer 1

Answer: 98

Let $z=x+iy$
$\arg \left(\frac{x-2+iy}{x+2+iy}\right)=\frac{\pi }{4}$
$\arg (x-2+iy)-\arg (x+2+iy)=\frac{\pi }{4}$
${\tan }^{-1}\left(\frac{y}{x-2}\right)-{\tan }^{-1}\left(\frac{y}{x+2}\right)=\frac{\pi }{4}$
$\frac{\frac{y}{x-2}-\frac{y}{x+2}}{1+\left(\frac{y}{x-2}\right)\cdot \left(\frac{y}{x+2}\right)}=\tan \frac{\pi }{4}=1$
$\frac{xy+2y-xy+2y}{x^2-4+y^2}=1$
$4y=x^2-4+y^2$
$x^2+y^2-4y-4=0$
locus is a circle with center $(0,2)$ & radius $=2\sqrt{2}$

min. value $=(AP{)}^2=(OP-OA{)}^2$
$=(9\sqrt{2}-2\sqrt{2}{)}^2$
$=(7\sqrt{2}{)}^2=98$