# How to determine shape of level curves of $V(y,z)= \ln{ \left( \frac{y^2 + (z+z_o)^2}{y^2 + (z-z_o)^2}\right)}$?

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?

• Take exponentials on both sides for constant values of $V$ and you will end up with the equation of a circle. Jan 9, 2021 at 17:30

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.