Question

If $z=x+iy$, then show that $z\overline{z}+2(z+\overline{z})+a=0$, where $a\in R$, represents a circle.

Answer 1

We have,

$z=x+iy$

$\overline{z}=x-iy$

Therefore, $z\overline{z}+2(z+\overline{z})+a=0$

$(x+iy)(x-iy)+2(x+iy+x-iy)+a=0$

$x^2+y^2+4x+a=0$

$(x+2{)}^2+(y-0{)}^2=(\sqrt{4-a}{)}^2$, which represents a circle for all $a\le 4$.