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I have the equation $$ x^2 + y^2 = 2\left(\sqrt{x^2 + y^2}\,\right) + 2y + 3x $$ and I want to solve it in terms of $y$ with no square roots.

I'm having trouble with proceeding from here. My initial thought was to isolate the square root, then square both sides of the equation, but I'm unsure whether that would actually work. Any advice?

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    $\begingroup$ Your idea will work if you just carry out the computation! :) $\endgroup$ – H. R. Nov 16 '15 at 12:48
  • $\begingroup$ No, because then he will get a mixed term with $x$ and $y$. $\endgroup$ – Jimmy R. Nov 16 '15 at 13:01
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Write $$\sqrt{x^2+y^2}=:r\geq0,\quad x:=r\cos\phi,\quad y:=r\sin\phi\ .$$ Then your equation says $$r^2 -r(2+3\cos\phi+2\sin\phi)=0\ .$$ This is solved by $r=0$, giving $(x,y)=(0,0)$, and by $$r=2+3\cos\phi+2\sin\phi\ .\tag{1}$$ Determine the $\phi$-interval $J\subset\ ]{-\pi},\pi[\ $ for which the right hand side of $(1)$ is $\geq0$. The solutions of your equation are then parametrized by $$\left.\eqalign{x(\phi)&=(2+3\cos\phi+2\sin\phi)\cos\phi\cr y(\phi)&=(2+3\cos\phi+2\sin\phi)\sin\phi\cr}\quad\right\}\qquad(\phi\in J)\ .$$ The following figure shows the relevant Mathematica output.

enter image description here

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