Why are both $|z-z_0| = r$ and $|z-z_0|^2=r^2$ equations of a circle? Why are both $|z-z_0| = r$ and $|z-z_0|^2=r^2$ equations of a circle?
Specifically, why would the latter one describe the same circle as the former one?
 A: Because all that happens when you square both sides of an equation is that you potentially get some additional solutions from the alternative sign for the square root. But here everything is positive in both equations and there is no question of spurious solutions when you square or square root.
A: At least provided that $r \geq 0$, the two equations are equivalent: Both sides of the left equation are positive, so passing between them (by squaring and taking a square root) does not alter the solution set. As for why one might prefer the second equation, one reason is that, when written out in terms of the real and imaginary components of $z$ and $z_0$ (denote $z = x + i y$ and $z_0 = x_0 + i y_0$ for $x, y, x_0, y_0$ real), the second equation is polynomial, and hence particularly easy to use in some applications: $$(x - x_0)^2 + (y - y_0)^2 = r^2 .$$
If we take a square root to recover the first equation so written, we get
$$\sqrt{(x - x_0)^2 + (y - y_0)^2} = r ,$$
which is an equivalent and perfectly good equation, but, on account of the radical, it is not polynomial.
