Let $x,y$ be two variables. Consider the following system:
$$ \begin{cases} x=a_0+b (1+y^2/x^2)^{-1/2} \\ y=a_1+b(1+x^2/y^2)^{-1/2}, \end{cases}$$
where $a_0,a_1,b$ are parameters.
I can solve this using numerical solvers, but it's very inefficient. The two equations appear very symmetrical and somewhat 'simple' in formulation, and I wonder if there might be a way to solve this (maybe even obtain an approximation) in some analytic manner. Any hints or directions will be appreciated.
Thanks.
$$\begin{cases} x=a_0+b (1+y^2/x^2)^{-1/2} \\ y=a_1+b(1+x^2/y^2)^{-1/2} \end{cases} \implies \begin{cases} (x-a_0)^2=\dfrac{b^2x^2}{x^2+y^2} \\ (y-a_1)^2=\dfrac{b^2y^2}{x^2+y^2} \end{cases}\implies \dfrac{(x-a_0)^2}{x^2}=\dfrac{(y-a_1)^2}{y^2}$$
$$\implies \left|\dfrac{x-a_0}{x}\right|= \left|\dfrac{y-a_1}{y}\right|\implies \dfrac{y-a_1}{y}=\pm \dfrac{x-a_0}{x}.$$ So, we get that $$y=\dfrac{a_1x}{a_0} \quad \mathrm{or}\quad y=\dfrac{a_1x}{2x-a_0}.$$
If $y=\dfrac{a_1x}{a_0}$ one gets $x:$ $$x=a_0+b (1+a_0^2/a_1^2)^{-1/2}$$ and $y=\dfrac{a_1}{a_0}(a_0+b (1+a_0^2/a_1^2)^{-1/2}).$
If $y=\dfrac{a_1x}{2x-a_0}$ then $\dfrac{(x-a_0)^2}{x^2}=\dfrac{(y-a_1)^2}{y^2}$ that $$a_1^2(x-a_0)^2=(a_1x+a_0)^2$$ is an identity. So, we consider
$$\dfrac{a_1x}{2x-a_0}=a_1+b\left(1+\dfrac{x^2(2x-a_0)^2}{a_1^2x^2}\right)^{-1/2}=a_1+b\left(1+\dfrac{(2x-a_0)^2}{a_1^2}\right)^{-1/2}.$$ Thus
$$\left(\dfrac{a_1x}{2x-a_0}-a_1\right)^2=\dfrac{b^2}{1+\dfrac{(2x-a_0)^2}{a_1^2}}=\dfrac{a_1^2b^2}{a_1^2+(2x-a_0)^2}.$$ It follows that
$$\left(\dfrac{a_0-x}{2x-a_0}\right)^2=\dfrac{b^2}{a_1^2+(2x-a_0)^2}.$$ From this we get that
$$a_1^2(a_0-x)^2=(2x-a_0)^2(b^2-(a_0-x)^2).$$
Since $a_1^2(a_0-x)^2\ge 0$ it must be $b^2\ge (a_0-x)^2$ (unless $2x=a_0$) But, from $x=a_0+b (1+y^2/x^2)^{-1/2} $ we have
$$(x-a_0)^2=\dfrac{b^2}{1+\dfrac{y^2}{x^2}}< b^2.$$ This is not possible. Now, if $2x=a_0$ then it must be $a_1^2(a_0-x)^2=0.$ That is, $a_1=0,$ from where $y=0.$ But, since we are taking division by $y$ in the original equation this is not possible.
In conclusion, the solutions of the system (when $a_0a_1\ne 0$) are given by
$$x=a_0+b (1+a_0^2/a_1^2)^{-1/2}, y=\dfrac{a_1}{a_0}(a_0+b (1+a_0^2/a_1^2)^{-1/2}).$$
Note that from $$\dfrac{(x-a_0)^2}{x^2}=\dfrac{(y-a_1)^2}{y^2}$$ we get that $a_0=0\iff a_1=0.$ In such a case it is $x=y=\sqrt{b^2-1}.$