Question

If $z=x +i y$ is a complex number satisfying $\left|\frac{z-2 i}{z+2 i}\right|=2$ and the locus of $z$ is a circle, then its radius is

• A. $\frac{5}{3}$
• B. $\sqrt{\frac{71}{9}}$
• C. $\frac{8}{3}$
• D. $\frac{1}{3}$

Answer 1

Answer: (C)

We have,
$\left|\frac{z-2 i}{z+2 i}\right|=2$
$\Rightarrow \left|\frac{x+i y-2 i}{x+i y+2 i}\right|=2$
$\Rightarrow \left|\frac{x+(y-2) i}{x+(y+2) i}\right|=2$
$\Rightarrow \frac{|x+(y-2) i|}{|x+(y+2) i|}=2$
$\Rightarrow \frac{\sqrt{x^{2}+(y-2)^{2}}}{\sqrt{x^{2}+(y+2)^{2}}}=2$
$\Rightarrow x^{2}+(y-2)^{2}=4\left[x^{2}+(y+2)^{2}\right]$
$\Rightarrow x^{2}+y^{2}-4 y+4=4 x^{2}+4 y^{2}+16 y+16$
$\Rightarrow 3 x^{2}+3 y^{2}+20 y+12=0$
$\Rightarrow x^{2}+y^{2}+\frac{20}{3} y+4=0$
$\Rightarrow x^{2}+\left(y+\frac{10}{3}\right)^{2}+4-\frac{100}{9}=0$
$\Rightarrow x^{2}+\left(y+\frac{10}{3}\right)^{2}=\left(\frac{8}{3}\right)^{2}$
Which is equation of circle with centre $\left(0, \frac{-10}{3}\right)$ and radius $\frac{8}{3}$.