The locus of a point $z$ satisfying $|z|^2=\operatorname{Re}(z)$ is a circle with centre
The locus of a point $z$ satisfying $|z|^2=\operatorname{Re}(z)$ is a circle with centre
$\left(0, \frac{1}{2}\right)$
$\left(-\frac{1}{2}, 0\right)$
$\left(\frac{1}{2}, 0\right)$
$\left(0,-\frac{1}{2}\right)$
Solution
Let $z=x+i y$
$|z|=\sqrt{x^2+y^2}$
Now, $|z|^2=\operatorname{Re}(z)$
$x^2+y^2=x$
$\Rightarrow \quad x^2+y^2-x=0$
$g=1 / 2, f=0$
So, centre of circle $\left(\frac{1}{2}, 0\right)$.