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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?

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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

2 Answers 2

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The line from the centre of the circle to the centre of the sphere will be perpendicular to the plane. To put it another way, the line perpendicular to the plane which passes through the centre of the sphere will intersect the plane at the centre of the circle. This allows you to calculate the centre of the circle.

You can then calculate the distance between the centre of the circle and the centre of the sphere. You already know (I hope) the distance between the centre of the sphere and the circle. An application of Pythagorus' Theorem will then give the distance between the centre of the circle and the circle, in other words the radius of the circle.

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Yes, of course! The center of sphere is at the origin, in which I can know take the distance between the origin and the plane! Thank you! –  Curtain Jan 29 '13 at 18:12

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.

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Thanks for your hint! –  Curtain Jan 29 '13 at 18:13

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