Equation of a circle in a complex plane

Solution 1:

A brute force approach : Write $z = x+iy, z_1 = x_1+iy_1, z_2 = x_2+iy_2$, and simplify the formula $$ |z-z_1|^2 = c^2|z-z_2|^2 $$ You will end up with an equation of the form $$ (1-c^2)x^2 + (1-c^2)y^2 + 2\alpha x + 2\beta y +\gamma = 0 $$ for some constants $\alpha, \beta, \gamma$. Since $c\neq 1$, you can divide by $(1-c^2)$ (Note that $c$ is a real number and $\neq -1$ because of what you started with). This will give you the equation of a circle.

Solution 2:

It is the same as $|z-z_1|^2=c^2|z-z_2|^2$.
Write $z=x+iy$, $z_1=x_1+iy_1$ and so on.
Expand the squared absolute values, and simplify.