<s>[INST] Find the intersection curve between a plane $x+y=1$ and a sphere $x^{2}+y^{2}+z^{2}=1$. Given two equations:
$\left\{\begin{matrix}
x^{2}+y^{2}+z^{2}=1\\ 
x+y=1
\end{matrix}\right.$
Find the set of points in $3$-space represented by this pair.
My work so far:
The equation $x^{2}+y^{2}+z^{2}=1$ represents a sphere with radius $1$ centered at the origin. The equation $x+y=1$ represents a plane in the $xy$-plane that has the points (1, 0, 0) and (0, 1, 0). 
I am not sure how to find the equation of the curve (the circle) that is formed from the intersection of these two equations. Do I use substitution?
Let $x = 1 - y$.
$(1-y)^{2}+y^{2}+z^{2}=1$
$1-2y+y^{2}+y^{2}+z^{2}=1$
$2\left (y-\frac{1}{2}  \right )^{2}+z^{2}=\frac{1}{2}$
I am close to the answer in the textbook, which is a circle with radius $\frac{1}{\sqrt{2}}$ centered at (1/2, 1/2, 0).
I know that the circle is in the $xy$-plane, so $z = 0$. However, how did they get the $x = \frac{1}{2}$ coordinate when I don't see $x$ in the equation...?
 [/INST] Points of intersection will satisfy both equations, thus $x^2+y^2+z^2=x+y=1$. Simplifying $x^2+y^2+z^2=x+y$ we get $x^2-x+y^2-y+z^2=0$ or $(x-\frac{1}{2})^2+(y-\frac{1}{2})^2+z^2=\frac{1}{2}$. Don't be alarmed that it looks like another sphere, there is no general equation for a circle in 3d, we can only write it on a particular plane, which is $x+y=1$ in this case.
</s>