Solving $x^2+y^2=9$, $\arctan\frac{y+2}{x+2} + \arctan\frac{y-2}{x+4} =2\arctan\frac{y}{x}$ without graphing? The system of equations: 

$$\begin{align}
x^2 + y^2 &= 9 \\[6pt]
\operatorname{arctan}\frac{y+2}{x+2} + \operatorname{arctan}\frac{y-2}{x+4} &=2\,\operatorname{arctan} \frac{y}{x}
\end{align}$$ 

I tried to interpret the second equation by setting $x$ and $y$ as the legs of a right triangle, but I'm still unable to solve this system of of equations geometrically. Any tips?

Essentially, the equations say that the origin is at distance $3$ from $(x,y)$ and on the angular bisector of the triangle with vertices $(x,y)$, $(x+2,y+2)$, $(x+4,y-2)$.
