Solving $n(x-\bar x)=\sum_{i=1}^nd_i{\frac{x-x_i}{|| (x,y)-(x_i,y_i)||}}$ and $n(y-\bar y)=\sum_{i=1}^nd_i{ \frac{y-y_i}{||(x,y)-(x_i,y_i)||}}$
In connection to this answer of mine, I got the equations $$nx-\sum_{i=1}^{n}x_i = \sum_{i=1}^{n} d_i\frac{x-x_i}{\sqrt{(x-x_i)^2+(y-y_i)^2}}$$ and $$ny-\sum_{i=1}^{n}y_i = \sum_{i=1}^{n} d_i\frac{y-y_i}{\sqrt{(x-x_i)^2+(y-y_i)^2}}$$ where the only unknowns are $x, y$.
I doubt it is possible to get a closed form solution for $x, y$ for $n \ge 3$, but I do think $x, y$ can be written in terms of roots of polynomials. For example, if $n = 2$, we get that $$(2x-x_1-x_2)^2 - \frac{(x_2-x_1)^2 (d_2-d_1)^2}{(x_2-x_1)^2 + (y_2-y_1)^2} = 0$$ and $$(2y-y_1-y_2)^2 - \frac{(y_2-y_1)^2 (d_2-d_1)^2}{(x_2-x_1)^2 + (y_2-y_1)^2} = 0$$
This can of course be solved explicitly for $x, y$. I got this result by squaring both sides. For higher $n$, I don't think squaring both sides will help, as there will be more square roots after squaring than before. This leads to my question: How can I get polynomials like these for higher $n$?
