<s>[INST] radius of a sphere containing a circle Consider the circle $C$ which is the intersection of the sphere $x^2+y^2+z^2-x-y-z=0$ and the plane $x+y+z=1$. The radius of the sphere with centre at the origin, containing the circle $C$ is 
a. $1$
b. $2$
c. $3$
d. $4$
 [/INST] When we complete the squares in the equation for the sphere, we find
$$ (x^2 - x + \frac{1}{4}) + (y^2 - y + \frac{1}{4}) + (z^2 - z + \frac{1}{4}) \ = \frac{3}{4} $$
$$ \Rightarrow \ (x - \frac{1}{2})^2 \ + \ (y - \frac{1}{2})^2 \ + \ (z - \frac{1}{2})^2 \ = \frac{3}{4}  \ \ . $$
An immediate way to find the circle would be to first consider the line from the origin through the center of the sphere, $ \ x = y = z = t \ . $  The center of the circle is also on this line: using the equation of the plane, we have $ \ t + t + t \ = \ 1 \ \Rightarrow \ t \ = \ \frac{1}{3} \ . $  The center of the circle is thus at $ \ (\frac{1}{3} , \frac{1}{3} , \frac{1}{3}) \ . $
Now, the distance from the center of the sphere to the center of this circle is $ \ s \ = \ \sqrt{(\frac{1}{2} - \frac{1}{3})^2 + (\frac{1}{2} - \frac{1}{3})^2 + (\frac{1}{2} - \frac{1}{3})^2} \ = \ \sqrt{\frac{1}{12}} \ . \ $ The radius of the circle is one leg of a right triangle with a hypotenuse equal to the radius of the sphere, $ \ R \ = \ \frac{\sqrt{3}}{2} \ $ and the other leg being $ \ s \ . \ $  So the radius of the circle is $ \ \sqrt{(\sqrt{\frac{3}{4}})^2 - (\sqrt{\frac{1}{12}})^2} \ = \ \sqrt{\frac{2}{3}} \ . \ $
There is also a right triangle formed by the radius of the circle, the segment from the origin to the center of the circle, and the radius of the origin-centered sphere, which is the hypotenuse.  The center of the circle is at a distance $  \ \sqrt{( \frac{1}{3})^2 + (\frac{1}{3})^2 + (\frac{1}{3})^2} \ = \ \sqrt{\frac{1}{3}} \  $ from the origin.  Thus the radius of our sphere is  $ \ \sqrt{(\sqrt{\frac{1}{3}})^2 + (\sqrt{\frac{2}{3}})^2} \ = \ 1 \ . \ \ $ [choice (a)]
(RicardoCruz got his illustration posted first; here's another view.)

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