$\Large \begin{align*}

&u = \frac{\sqrt{4 r^2-b^2}}{b}+2 y \\

\\

&v = \frac{2 b \sqrt{4 r^2-b^2} (x-y)+b^2-4 r^2}{b^2-4 r^2}\\

\\

&a = 2 \left(\tan^{-1}\left(-\frac{\sqrt{4 r^2-b^2}}{r},\frac{b}{r}\right)\right)

\end{align*}$

So, I tried plugging these into f(u,v,a) but the result is not the circle x,y,r. In other words, it doesn't work. I rewrote the formulas 2 times, always with the same result. I also tried to figure the inverse out myself and think I've got it right ideawise:

I had "a" figured out for "r" already, so I just plugged it into "x" and "y":

\(\displaystyle

a = 2*sin^{-1}(b/2*r) \\

\)

\(\displaystyle

x = \frac{u}{2} + v* \frac{1+cos(a)}{2*sin(a)} \\

\)

\(\displaystyle

x = \frac{u}{2} + v* \frac{1+cos(2*sin^{-1}(b/2*r))}{2*sin(2*sin^{-1}(b/2*r))} \\

\)

\(\displaystyle

u = \frac{x}{2} + v*\frac{1 \pm sqrt(1-(b/2*r)^2)}{-2b/r}

\)

\(\displaystyle

y = \frac{v}{2} + u* \frac{1+cos(a)}{2*sin(a)} \\

\)

\(\displaystyle

v = \frac{y}{2} + u*\frac{1 \pm sqrt(1-(b/2*r)^2)}{-2b/r}

\)

Now I plugged "v" into "u":

\(\displaystyle

u = \frac{x}{2} + (\frac{y}{2} + u*\frac{1 \pm sqrt(1-(b/2*r)^2)}{-2b/r} )*\frac{1 \pm sqrt(1-(b/2*r)^2)}{-2b/r}

\)

Now I just need to get all the u's to the left and I've got the formula for u.

Is that correct? I did this already, but the result was also not working, although it almost works, which is, why I think there's probably a little mistake in the simplification-process, let me show you:

The green circle is fed directly by the parameters x,y,r and the red one goes through the inverse of f and f itself.

If you find any mistake, let me know