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?

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Find the curve of intersection between $\ds{x^{2} + y^{2} + z^{2} = 1}$ and $\ds{x + y + z = 0}$.

\begin{align}
1 & = x^{2} + y^{2} + \pars{-x - y}^{2} =
2x^{2} + 2y^{2} + 2xy
\\[5mm] & =
2\bracks{{x + y \over 2} + {x - y \over 2}}^{2} +
2\bracks{{x + y \over 2} - {x - y \over 2}}^{2} + 
2\bracks{{x + y \over 2} + {x - y \over 2}}\bracks{{x + y \over 2} - {x - y \over 2}}
\\[5mm] & =
{3 \over 2}\pars{x + y}^{2} + {1 \over 2}\pars{x - y}^{2}.
\\ & \text{which is}\ identically\ satisfied\ \text{by}\qquad
\left\{\begin{array}{rcrcr}
\ds{x} & \ds{+} & \ds{y} & \ds{=} & \ds{{\root{6} \over 3}\cos\pars{\theta}}
\\
\ds{x} & \ds{-} & \ds{y} & \ds{=} & \ds{\root{2}\sin\pars{\theta}}
\end{array}\right.
\end{align}

$$
\bbx{x =
{\root{6} \over 6}\,\cos\pars{\theta} +
{\root{2} \over 2}\,\sin\pars{\theta}\,,\qquad
y = {\root{6} \over 6}\,\cos\pars{\theta} -
{\root{2} \over 2}\,\sin\pars{\theta}}
$$
