# Calc III: sphere & plane intersection.

Consider the intersection of the sphere (x-3)^2+(y+2)^2+(z-1)^2=13 with the plane x+y=0

a) This intersection should be a familiar curve. Describe the curve.

b) Substituing x=-y or y=-x into the equation of the sphere does not give this curve. Explain the difference.

I know that the intersectionis a circle but I am having trouble proving that. I rearranged the above equation and set it equal to x+y:

(x-3)^2+(y+2)^2+(z-1)^2-13=x+y

The next thing I did was put the like terms together to give the following equation:

x^2-6x+9-x+y^2+4y+4+y+z^2-2z+1-13=0

I then simplified this equation:

X^2-7x+9+y^2+3y+4+(z-1)^2-13=0

Now I am stuck. This just looks like a smaller sphere.
Also, I would plug in x=-y or vice versa but the second part of the question says that will not give me the result that I want.

## The Attempt at a Solution

Mark44
Mentor
The intersection of a sphere and a plane will be one of the following.
1. The empty set (the two don't actually intersect).
2. A single point.
3. A circle.

Last edited:
tiny-tim
Homework Helper
Hi hq1330! (try using the X2 tag just above the Reply box )
Consider the intersection of the sphere (x-3)^2+(y+2)^2+(z-1)^2=13 with the plane x+y=0

a) This intersection should be a familiar curve. Describe the curve.

b) Substituing x=-y or y=-x into the equation of the sphere does not give this curve. Explain the difference.

Also, I would plug in x=-y or vice versa but the second part of the question says that will not give me the result that I want.

I think this question is unnecessarily obscure. A curve, in three dimensions, requires two equations to describe it, so of course substituing x=-y or y=-x into one equation, which will only give one equation, cannot give the curve.

(in fact, it will give the two-dimensional projection of the curve onto the xz or yz plane)

Sooo … plug away! (also, the method you used gives a whole surface, including eg the intersection of the sphere (x-3)2+(y+2)2+(z-1)2=14 with the plane x+y=1, doesn't it? )