Expert: Alon Mandes - 3/29/2009

Question
QUESTION: The intersection of a plane and a sphere is a circle.  So how do you obtain the equation of the circle from intersecting x^2+y^2+z^2 = 4 with the plane y=x?  I get an equation of an ellipse.  Do you have to consider the angle that the plane y=x makes with the yz-plane?

ANSWER: This angle won't make a diffrence. The plane y=x will cut an ellipse.
The only planes that cut a circle by intersection with the sphere
are : x=a , y=b , z=c .

Alon.

---------- FOLLOW-UP ----------

QUESTION: I have to disagree.  Think about it.  Obviously one can show algebraically that the planes  x=a , y=b , z=c intersect the sphere in circles.  But I have to repeat, "the intersection of any intersecting plane and a sphere is a circle".  The plane y=x intersects the sphere, and it has to be a circle.  My question is how do you come up with the equation of that circle in that plane?

Answer
I have to disagree with you. The  intersection of any intersecting plane and a sphere is an ellipse. & when the planes are canonic
from the form I wrote above then the ellipse becomes circle.
Here is an example :
The plane is : y=x,
so x^2+x^2+z^2=4 -> 2x^2+z^2=4 -> x^2/2 + y^2/4 = 1 Ellipse.
Lets ake it more further :
The plane is ax+by=1 ,
so x^2 + (1-ax)^2/b^2 + z^2 = 4
x^2 + (1-2ax+a^2x^2)/b^2 + z^2 = 4
x^2 + [(a/b)^2]x^2 -(2a/b^2)x + z^2 = 4-1/b^2
& this is definitely an ellipse.

Alon.