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

Given: $\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi }{4}$
So, $z$ lies on an arc of a circle as shown in figure.
Centre of this circle is $(0,2)$ and radius $=2\sqrt{2}$
Now,
${|z-9\sqrt{2}-2i|}^2=|z-(9\sqrt{2}+2i\left)\right|$
$=$ Distance of $z$ from $(9\sqrt{2},2)$
Distance of $(9\sqrt{2},2)$ from centre (0,2)
$=9\sqrt{2}$
Minimum value of $|z-9\sqrt{2}-2i{|}^2$
$={(9\sqrt{2}-2\sqrt{2})}^2=98$