Discussion Forum

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

Answer:

Explanation:

 We have  

$|\frac{z-2i}{z+2i}|=2 \Rightarrow$ $|\frac{x+iy-2i}{x+iy+2i}|=2$

 $\Rightarrow$     $\mid\frac{x+(y-2)i}{x+(y+2)i}|=2\Rightarrow\frac{|x+(y-2)i|}{|x+(y+2)i|}=2$

$\Rightarrow$    $\frac{\sqrt{x^{2}+(y-2)^{2}}}{\sqrt{x^{2}+(y+2)^{2}}}=2$

$\Rightarrow$          $x^{2}+(y-2)^{2}=4[x^{2}+(y+2)^{2}]$

$\Rightarrow$            $x^{2}+y^{2}-4y+4=4x^{2}+4y^{2}+16y+16$

 $\Rightarrow$       $x^{2}+y^{2}+\frac{20}{3}y+4=0$

$\Rightarrow$      $x^{2}+\left(y+\frac{10}{3}\right)^{2}+4-\frac{100}{9}=0$

 $\Rightarrow$    $x^{2}+\left(y+\frac{10}{3}\right)^{2}=\left(\frac{8}{3}\right)^{2}$

 Which is equation of circle with centre $(0, \frac{-10}{3})$ and radius $\frac{8}{3}$