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...?
 A: 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 $R=\frac{1}{\sqrt2}$.
A: You know from geometry that the intersection is a circle, but the writing the equation of a circle in 3D isn't so straightforward (different forms can exist).
To ease the task, we can rotate space in such a way that the plane becomes parallel to $yz$. We will use the transform
$$x,y,z\to\frac{x+y}{\sqrt2},\frac{x-y}{\sqrt2},z.$$
The equation of the plane becomes
$$x=\frac1{\sqrt2}$$ and that of the sphere
$$x^2+y^2+z^2=1,$$ which is no surprise as the sphere is symmetrical.
Now the circle has the 2D equation 
$$y^2+z^2=\frac12.$$
It has the radius $\dfrac1{\sqrt2}$ and the center $\left(\dfrac1{\sqrt2},0,0\right)$ in the rotated coordinates. We counter-rotate with
$$x,y,z\to\frac{x+y}{\sqrt2},\frac{x-y}{\sqrt2},z$$
and the center moves to
$$\left(\frac12,\frac12,0\right).$$
A: 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.
