Equation of the complex locus: $|z-1|=2|z +1|$ 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 $(-1,0)$?

Just to add on to Aryabhata's comment above. The map $f(z) = \frac{1}{z}$ for $ z \in \mathbb{C} -\{0\}$, $f(0) = \infty$ and $f(\infty) = 0$ is a circle preserving homeomorphism of $\bar{\mathbb{C}}$. To see this, one needs to prove that it is continuous on $\bar{\mathbb{C}}$, and since $f(z)$ is an involution proving this would mean that its inverse is continuous as well. It is also not hard to show that $f(z)$ is bijective.
Lastly use the general equation of a circle in $\bar{\mathbb{C}}$ to see that circles in $\bar{\mathbb{C}}$ are preserved under this map.
