The locus of the point $z=x+i y$ satisfying $\left|\frac{z-2 i}{z+2 i}\right|=1$ is

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Given, }\left|\frac{z-2 i}{z+2 i}ight|=1 \ & \Rightarrow \quad \frac{|x+i y-2 i|}{|x+i y+2 i|}=1\end{aligned}$ $\begin{aligned} & \Rightarrow \sqrt{x^2+(y-2)^2}=\sqrt{x^2+(y+2)^2} \ & \Rightarrow x^2+y^2+4-4 y=x^2+y^2+4+4 y \ & \Rightarrow \quad 8 y=0 \ & \Rightarrow \quad y=0 \quad 	ext { ie, } x 	ext {-axis. } \ & \end{aligned}$

The locus of the point $z=x+i y$ satisfying $\left|\frac{z-2 i}{z+2 i}\right|=1$ is

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