Question

If $z=x+iy$ and $|z-1{|}^2+|z+1{|}^2=4$, then the locus of $z$ is

• A. $x^2-y^2=1$
• B. $x^2+y^2=1$
• C. $x^2-y^2=4$
• D. $x^2+y^2=2$

Answer 1

The correct option is B $x^2+y^2=1$

Let $z=x+iy$
$\Rightarrow |x+iy-1{|}^2+|x+iy+1{|}^2=4\Rightarrow |x-1+iy{|}^2+|x+1+iy{|}^2=4\Rightarrow (x-1{)}^2+y^2+(x+1{)}^2+y^2=4\Rightarrow 2x^2+2y^2=2$

Hence, the required locus is $x^2+y^2=1$