Intersection curve between a circle and a plane - Stokes theorem What is the intersection curve between the circle
$$x^2+y^2=1$$
and the plane
$$x+y+z=0$$
If i am not wrong, I should solve the equation system
\begin{align}
x^2+y^2-1=0 \\
x+y+z=0
\end{align}
But I don't get the right curve. If i solve x in the first equation, then i get the system\begin{align}
x=- \sqrt{1-y^2} \\
x+y+z=0
\end{align}
and
\begin{align}
x= \sqrt{1-y^2} \\
x+y+z=0
\end{align}.
And this gives me
$$x=- \sqrt{1-y^2} => z= -y + \sqrt{1-y^2}$$
$$x=\sqrt{1-y^2} => z= -y - \sqrt{1-y^2}$$But it's wrong. Where am i doing wrong?
EDIT: If I have to calculate the work of a field are doing along the curve p, do I need to calculate the curve or should I use the plane equation to get the normal vector??
That's what this guy here
Stokes's Theorem on a Curve of Intersection did. But I just wan't to know if I am always going to do it so  I don't make any mistakes. And i would appreciate if you could link any good sites to learn more about this, thanks!
If i do parametrize the plane equation as
\begin{align}
x=s \\
y=t \\
z=-s-t \\
\end{align}
then i will get the normal vector
$$r_s x  r_t =(1,1,1)$$
and with this I can use stokes theorem with x^2+y^2<=1, right?

I think the intersection just two points.
$$
(\sqrt{2}/2, -\sqrt{2}/2, 0)
$$ 
and
$$
(-\sqrt{2}/2, \sqrt{2}/2, 0).
$$
