Solve the following


If $\left|\frac{z-2}{z+2}\right|=\frac{\pi}{6}$, then the locus of $z$ is


Given $\left|\frac{z-2}{z+2}\right|=\frac{\pi}{6}$

i. e $|z-2|=\frac{\pi}{6}|z+2|$

i. e $|z-2|^{2}=\frac{\pi^{2}}{6^{2}}|z+2|^{2} \quad$ for $z=x+i y$

i. e $|x+i y-2|^{2}=\frac{\pi^{2}}{36}|x+i y+2|^{2}$

i. e $|(x-2)+i y|^{2}=\frac{\pi^{2}}{36}|(x+2)+i y|^{2}$

i. e $(x-2)^{2}+y^{2}=\frac{\pi^{2}}{36}\left[(x+2)^{2}+y^{2}\right]$

i. e $x^{2}+y^{2}+4-4 x=\frac{\pi^{2}}{36}\left[x^{2}+y^{2}+4 x+4\right]$

i.e $\left(1-\frac{\pi^{2}}{36}\right) x^{2}+\left(1-\frac{\pi^{2}}{36}\right) y^{2}+4\left(1-\frac{\pi^{2}}{36}\right)-4 x\left(1+\frac{\pi^{2}}{36}\right)=0$

which defines locus of a  circle.

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