<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] Your derivation is correct but from here
$$2\left (y-\frac{1}{2}  \right )^{2}+z^{2}=\frac{1}{2}$$
$$y = 1 - x$$
you find
$$\left (1-x-\frac{1}{2}  \right )^{2}+\left (y-\frac{1}{2}  \right )^{2}+z^{2}=\frac{1}{2}$$
$$\left (x-\frac{1}{2}  \right )^{2}+\left (y-\frac{1}{2}  \right )^{2}+z^{2}=\frac{1}{2}$$
which is the equation of a sphere centered in $C(\frac12,\frac12,0)$ and with radius $R=\frac{1}{\sqrt2}$.
Since the center belongs to the plane $x+y=1$ then the intersection between the sphere and the plane is a circle with center $C(\frac12,\frac12,0)$ and radius $R=\frac{1}{\sqrt2}$.
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