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.


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?

enter image description here

  • 1
    $\begingroup$ 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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.