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?

$2x^2+2xy+2y^2=1$ can be rewritten as $(2x+y)^2+3y^2=2$ and  this  can be parameterised 
\begin{eqnarray*}
\frac{2x+y}{\sqrt{2}} &=& \cos \theta  \\
\sqrt{\frac{3}{2}} y &=& \sin \theta . \\
\end{eqnarray*}
now substitute these into $x+y+z=0$ and we have 
\begin{eqnarray*}
x&=&\frac{1}{\sqrt{2}} \cos \theta- \frac{1}{\sqrt{6} } \sin \theta \\
y &=&\sqrt{ \frac{2}{3}} \sin \theta \\
z&=& - \frac{1}{\sqrt{2}} \cos \theta - \frac{1}{\sqrt{6}} \sin \theta.
\end{eqnarray*}
Alternatively 
\begin{eqnarray*}
\left(\begin{array}{c}
x  \\ y \\ z
\end{array}\right)=\left(\begin{array}{c}
\frac{1}{\sqrt{2}}  \\ 0 \\ -\frac{1}{\sqrt{2}}
\end{array}\right) \cos \theta+\left(\begin{array}{c}
-\frac{1}{\sqrt{6}}  \\ \sqrt{\frac{2}{3}} \\ -\frac{1}{\sqrt{6}}
\end{array}\right) \sin \theta
\end{eqnarray*}
now observe that these are orthogonal unit vectors and so this is a circle.
