Find the cut of the sphere and the given plane I'm trying to solve an integral from my homework, however, to do that, I need to find the circle cut of the next sphere and plane:

$x^2 +y^2 +z^2 = 1 $  
$x + y + z = 1 $
I found the cutting points with the x,y,z axis, but didn't manage to find the radius or the center of the circle cut.
I would appreciate any help, Thanks.
 A: Step 1: From the plane, $\;z=1-(x+y)\;$ ;
Step 2: Substitute the above in sphere :
$$x^2+y^2+\left(1-(x+y)\right)^2=1\implies x^2+y^2-x-y+xy=0\;\;\color{red}{(*)}$$
Step 3: Knowing that we  always get a circle as intersection if it is not empty or only one point, we try to parametrize Step 2 as a circle, for example by means of completing the square:
$$\color{red}{(*)}\;\;\left(x+\frac{y-1}2\right)^2+\frac34y^2-\frac y2-\frac14=0\implies\left(x+\frac{y-1}2\right)^2+\frac34\left(y-\frac13\right)^2=\frac13\;\;\color{blue}{(**)}$$$${}$$
Step 4:  "Force" a circle parametrization here, by means of:
$$x':=\left(x+\frac{y-1}2\right)\;,\;\;y':=\frac{\sqrt3}2\left(y-\frac13\right)\implies\color{blue}{(**)}\;\;x'^2+y'^2=\left(\frac1{\sqrt3}\right)^2$$
or with polar parametrization:
$$\begin{cases}
x+\cfrac{y-1}2=\cfrac1{\sqrt3}\cos\theta\implies x+\cfrac12y=\cfrac1{\sqrt3}\cos\theta+\cfrac12\\{}\\
\cfrac{\sqrt3}2\left(y-\cfrac13\right)=\cfrac1{\sqrt3}\sin\theta\implies\cfrac12y=\cfrac13\sin\theta+\cfrac16\end{cases}$$
and substracting we get
$$\begin{cases}x=\cfrac13\left(\sqrt3\,\cos\theta-\sin\theta\right)+\cfrac13\\{}\\
y=\cfrac23\sin\theta+\cfrac13\end{cases}$$
and finally the last coordinate is
$$z=1-x-y=1-\cfrac13\left(\sqrt3\,\cos\theta-\sin\theta\right)-\cfrac13-\cfrac23\sin\theta-\cfrac13=$$
$$=\frac13-\frac13\left(\sqrt3\,\cos\theta+\sin\theta\right)$$
and our circle is
$$r(\theta)=\left(\,\cfrac13\left(\sqrt3\,\cos\theta-\sin\theta\right)+\cfrac13\,,\,\,\cfrac23\sin\theta+\cfrac13\,,\,\,\frac13-\frac13\left(\sqrt3\,\cos\theta+\sin\theta\right)\,\right)$$
Please do check the above is a circle with center $\;\left(\cfrac13,\,\cfrac13,\,\cfrac13\right)\;$ (this should have been expected. Why?),   and radius $\;\sqrt\frac23\;$
