Question

Find the locus of the complex number, Z = x + iy
given $\left|\frac{x+iy-2i}{x+iy+2i}\right|=\sqrt{2}$

Answer 1

Given, $\left|\frac{x+iy-2i}{x+iy+2i}\right|=\sqrt{2}$
$\Rightarrow \frac{|x+i(y-2)|}{|x+i(y+2)|}=\sqrt{2}$
$\Rightarrow \frac{\sqrt{x^2+(y-2{)}^2}}{\sqrt{x^2+(y+2{)}^2}}=\sqrt{2}$
Squaring both sides, we get
$x^2+(y-2{)}^2=2\{x^2+(y+2{)}^2\}$
$\Rightarrow x^2+y^2-4y+4=2x^2+2y^2+8y+8$
$\Rightarrow x^2+y^2+12y+4=0$
$\Rightarrow x^2+y^2+12y+4+32=32$
$\Rightarrow x^2+y^2+12y+36=32$
$\Rightarrow x^2+(y+6{)}^2=32$
$\Rightarrow (x-0{)}^2+\{y-(-6){\}}^2=(4\sqrt{2}{)}^2$
Which represents a circle with centre (0, -6) and radius $4\sqrt{2}$.