This question requires finding the Cartesian equation for the locus:
$|z-1| = 2|z+1|$
that is, where the modulus of $z -1$ is twice the modulus of $z+1$
I've solved this problem algebraically (by letting $z=x+iy$) as follows:
$\sqrt{(x-1)^2 + y^2} = 2\sqrt{(x+1)^2 + y^2}$
$(x-1)^2 + y^2 = 4\big((x+1)^2 + y^2\big)$
$x^2 - 2x + 1 + y^2 = 4x^2 + 8x + 4 + 4y^2$
$3x^2 + 10x + 3y^2 = -3$
$x^2 + \frac{10}{3}x + y^2 = -1$
$(x + \frac{5}{3})^2 +y^2 = -1 + \frac{25}{9}$
therefore, $(x+\frac{5}{3})^2 + y^2 = \frac{16}{9}$, which is a circle.
However, I was wondering if there is a method, simply by inspection, of immediately concluding that the locus is a circle, based on some relation between the distance from $z$ to $(1,0)$ on the plane being twice the distance from $z$ to $(-1,0)$?