Consider this picture
which shows a coordinate system $(a,\tau)$ on the Cartesian coordinate frame $(x,y)$, which is very similar to the bipolar coordinate system, with isosurfaces (i.e. circles) $\tau$ and their respective foci $(-a,0)$ and $(a,0)$.
Can anyone help me derive formulae on how I might be able to go from Cartesian coordinates $(x,y)\mapsto (a,\tau)$ or polar coordinates $(\rho,\theta)\mapsto (a,\tau)$, similar to what you might find on https://en.wikipedia.org/wiki/Bipolar_coordinates?
Essentially, this problem arose because I'd like to sample a function of the form $f(\rho,\theta)=g(C\rho(\cos\theta,\sin\theta-1))$.
We have one family of curves $(x-a)^2+y^2=a^2$. Calculating the partial derivatives \begin{eqnarray*} \left( \frac{\partial a}{\partial x},\frac{\partial a}{\partial y} \right) = \left( \frac{x^2-y^2 }{2 x^2},\frac{y}{x} \right) \end{eqnarray*} We need a vector that is orthogonal to this \begin{eqnarray*} \left( \frac{\partial a}{\partial x},\frac{\partial a}{\partial y} \right) \cdot \left( \frac{\partial b}{\partial x},\frac{\partial b}{\partial y} \right) =0 \end{eqnarray*} So we need to solve the differential equation \begin{eqnarray*} (x^2-y^2) \frac{\partial b}{\partial x}+ 2xy \frac{\partial b}{\partial y} =0 \end{eqnarray*} Which is equivalent to solving \begin{eqnarray*} \frac{dy}{dx} =\frac{ (x^2-y^2)}{ 2xy } \end{eqnarray*} Let $y=ux$ and after a little calculus & algebra (& neat choice of arbitary constant) \begin{eqnarray*} x =\frac{2 bu}{ (1+u^2) } \end{eqnarray*} So \begin{eqnarray*} x^2 +(y-b)^2=b^2 \end{eqnarray*} So the other family of curves comes out to be circles as well ! ... but this time their centers run up the y-axis. \begin{eqnarray*} x^2 +(y-b)^2=b^2 \end{eqnarray*} To summarise ... \begin{eqnarray*} x & =& \frac{a b^2}{a^2+b^2} \ a & =& \frac{x^2+y^2}{2x} \\ y & =& \frac{a^2 b}{a^2+b^2} \ b & =& \frac{x^2+y^2}{2y} . \end{eqnarray*}