Context:
I expect that a circle will be written as $$R^2 = (x-a)^2 + (y-b)^2,$$ where $R$ is the radius, and $(a,b)$ is the center of the circle.
Question:
How to determine analytically that the level lines of $$V(y,z)= \ln{ \left( \frac{y^2 + (z+z_o)^2}{y^2 + (z-z_o)^2}\right)}$$ are circles?
Isocontours of $V$ are represented by constant values of the function, i.e., $V=C$, which yields
$$C = \ln{ \left( \frac{y^2 + (z+z_o)^2}{y^2 + (z-z_o)^2}\right)} \implies \mathrm{e}^C = \left( \frac{y^2 + (z+z_o)^2}{y^2 + (z-z_o)^2}\right)$$
Now rearrange to have
$$A \left[y^2 + (z-z_o)^2\right] = y^2 + (z+z_o)^2 $$
or, equivalently
$$ (A-1) y^2 + A(z-z_o)^2 - (z+z_o)^2 = 0,$$
with $A = \mathrm{e}^C $. Upon completing the square*, one has (if I didn't mess up)
$$ y^2 + (z-B)^2 = \frac{D}{A-1}, $$
where $B = z_o(A+1)/(A-1)$, and $D = 4Az_o^2/(A-1)$.
*Credit to this step goes to this answer.