Coordinate system $(a,\tau)$ for butterfly circles? 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*}
