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 $(-1,0)$?

View the point $z$ as a vector in $\mathbb R^2$. Without doing any calculations, we know that $\lVert z - (1,0) \rVert^2$ and $\lVert z - (-1,0) \rVert^2$ are of the form $z^Tz + b^T z + c$ for some $b$ and $c$. So the equation of the locus is $$(z^Tz + b_1^Tz + c_1) = 4(z^Tz + b_2^Tz + c_2),$$ or $$3z^Tz + b_3^Tz + c_3 = 0,$$ which is the equation of a circle.
In general, the form $x^TAx + b^Tx + c$ defines a quadric, and if you know something about the properties of the matrix $A$, it can tell you what the shape of the quadric is. In this case, $A$ is a multiple of the identity, so it's a circle/sphere/hypersphere.
