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?

Question

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?

Answer 1

Hint:

$$\arctan\dfrac{y+2}{x+2}-\arctan\dfrac yx=\arctan\dfrac yx-\arctan\dfrac{y-2}{x+4}$$

$$\iff\dfrac{2x-2y}{x^2+y^2+2x+2y}=\dfrac{2x+4y}{x^2+y^2-2x+4y}$$

As $x^2+y^2=9$ $$\iff\dfrac{2x-2y}{9+2x+2y}=\dfrac{2x+4y}{9-2x+4y}$$

$$\iff8x^2+16y^2+54y=0\ \ \ \ (1)$$

As $x^2+y^2=9,$ WLOG $x=3\cos t,y=3\sin t$

Put these values in $(1)$ and divide both sides by $\cos^2 t$

Answer 2

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