Find the curve of intersection between $x^2 + y^2 + z^2 = 1$ and $x+y+z = 0$ 
Find the curve of intersection between $x^2 + y^2 + z^2 = 1$ and $x+y+z = 0$

My attempt:


*

*$x^2 + y^2 + z^2 = 1$

*$x+y+z=0$


$$(2) \implies z = -(x+y)$$
$$(1) \implies x^2+y^2+(x+y)^2 = 1$$
$$2x^2 + 2y^2 + 2xy = 1$$
This is the curve in the xy-plane. Now if I could get y as a function of x, I could easily parametrize the curve but I am not able to do that. 
Is there an easier way to solve the problem?

HINT: it is $$y_{1,2}=-\frac{x}{2}\pm\sqrt{\frac{1}{2}-\frac{3}{4}x^2}$$ and with that $y$ we can compue $z$
