Simple solution to this algebraic equation?

Question

Consider this system of equations with two variables $x,y$ and all positive parameters $$\frac{2(x-w_1)}{a_1}=\frac{x}{\sqrt{x^2+y^2}}\,,\quad\frac{2(y-w_2)}{a_2}=\frac{y}{\sqrt{x^2+y^2}}.$$ It at most amounts to an algebraic equation of degree 4, which is always solvable in principle. But is there any possibility to explicitly solve this one? Hopefully some not too complicated form of solution? Thanks.

Answer 1

It follows from \begin{equation} \frac{(x-w_1)^2}{a_1^2} + \frac{(y-w_2)^2}{a_2^2} = \frac{1}{2^2} \end{equation} that the solutions are lying on the ellipses.

Thus \begin{equation} y = w_2 \pm \sqrt{\frac{a_2^2}{2^2} - \frac{a_2^2(x-w_1)^2}{a_1^2}} \end{equation} Then plugging this into the below \begin{equation} \frac{4}{a_1^2}(x-w_1)^2(x^2+y^2) = x^2 \end{equation} we have \begin{equation} -x^2 + \frac{4}{a_1^2}(x-w_1)^2\left(x^2 + \left[w_2 \pm \sqrt{\frac{a_2^2}{2^2} - \frac{a_2^2(x-w_1)^2}{a_1^2}} \right]^2\right) = 0 \end{equation} which is \begin{equation} -x^2 + \frac{4}{a_1^2}(x-w_1)^2\left(x^2 + w_2^2 + \frac{a_2^2}{2^2} - \frac{a_2^2(x-w_1)^2}{a_1^2} \pm 2w_2\sqrt{\frac{a_2^2}{2^2} - \frac{a_2^2(x-w_1)^2}{a_1^2}} \right) = 0 \end{equation}

Answer 2

$$\left(\frac{2(x-w_1)}{a_1}\right)^2+\left(\frac{2(y-w_2)}{a_2}\right)^2=\left(\frac{y}{\sqrt{x^2+y^2}}\right)^2+\left(\frac{x}{\sqrt{x^2+y^2}}\right)^2=1 \\ \implies \left(\frac{2(x-w_1)}{a_1}\right)^2+\left(\frac{2(y-w_2)}{a_2}\right)^2=1$$

This is an ellipse. Solutions are $$x-w_1=\frac{a_1}{2}\cos\theta \\ y-w_2 =\frac{a_2}{2}\sin\theta\quad \forall \quad\theta\in[0,2\pi]$$