# Sphere intersecting a plance

I have sphere and plane that intersects:

$$(1): x^2+y^2+z^2=2^2=4$$ $$(2): x+y+z=1$$

I know how to sketch this graph. I also know the resulting area. But how should I be able to get the center and the radius of circle?

I have tried to square complementing and subtract each equation, and to see if I can figure these things out. I have also tried to express eq. 2 as a function of (x,y) and put that in eq. 1.

Does someone know how to get the center point, and the radius?

-
What do you mean get the center and radius? Does this help: en.wikipedia.org/wiki/Sphere#Equations_in_R3 ? –  Git Gud Jan 29 '13 at 17:38
When they intersect, the resulting surface is a circle. I want to calculate the circle's center and radius. –  Curtain Jan 29 '13 at 17:42
Ooooooooooooohh, OK. –  Git Gud Jan 29 '13 at 17:43
@AviSteiner: That's confusing. The (book) answer says a circle. But that is in 2D right? Not all axes are there? –  Curtain Jan 29 '13 at 18:27
That's because my above comment was wrong. It's an ellipse when projected to the $xy$-plane. Sorry! –  Avi Steiner Jan 29 '13 at 18:28

The projection of the center of the sphere on the plane has coordinates $\frac{\sqrt 3}{3}(1,1,1)$ and it is the center of the circle. To calculate this, remember that the projection of the top of a pyramide on its base is the center of the triangle ; this center is situated at 2/3 of the altitude. Then apply Pythagore two times.