Question

you are given the sphere x^2+y^2+z^2=49. The plane x+y+z=1 intersect the sphere. Find the center and radius of the circle of intersection.

Answer 1

the radius is sqrt of 49-maginitude of normal vector of plane?

Answer 2

x^2+y^2+z^2=49
x+y+z=1
Then you need points that satisfy both equations, then x, y, z must be the same.
(x+y+z)49=49 (multiply by 49)
x^2+y^2+z^2=49(x+y+z)
x^2-49x+y^2-49y+z^2-49z=0 (sum 3*49^2/4)
(x^2-49x+49^2/4)+(y^2-49y+49^2/4)+(z^2-49z+49^2/4)=3/4*49^2
(x-24.5)^2+(y-24.5)^2+(z-24.5)^2=3/4*49^2.
Therefore, the center is (24.5, 24.5, 24.5)
and the radius is: 3/4*49^2

Answer 3

thanks