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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?

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    $\begingroup$ Take exponentials on both sides for constant values of $V$ and you will end up with the equation of a circle. $\endgroup$
    – Dmoreno
    Jan 9, 2021 at 17:30
  • $\begingroup$ Please see my answer $\endgroup$
    – Dmoreno
    Jan 9, 2021 at 18:12

1 Answer 1

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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.

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