Equation of a circle in a complex plane In the book I'm reading, one of the exercises starts with mentioning that $\left|\frac{z-z_1}{z-z_2}\right|=c$, where $c\neq 1$ is a constant, is an equation of the circle. 

It seems clear to me that $|z-z_1|=c$ is an equation of the circle in a complex plane but I can't figure out why the previous formula would also be an equation of the circle. Help with understanding would be appreciated.
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.
From an understanding point of view, if $ |z−z_1|=c  $ is  a circle. Then $  {∣z−z_1|\over|z−z_2∣}=c$, where c≠1 can be written as $|z−z_1|=c|z−z_2|$ which clearly looks like $ |z−z_1|$ is a circle scaled by a factor of c from another circle $ |z−z_2|=k$.

Hope that makes it clearer.