Intersection of a sphere and plane

[-EquinoX-]

Homework Statement

Say I have a sphere with the equation:

(x-4)^2 + (y-3)^2 + (z+2)^2 = 20

How do I find the intersection with the xy-plane? Is the intersection at a point or it forms a circle

Homework Equations

The Attempt at a Solution

Hmm.. I tried to set z equal to 0 to solve this problem, however I think that's not the way to solve this problem

 

[Dick]

If you are in the xy plane, then z=0. What's wrong with setting z=0?

 

[-EquinoX-]

then if I am asked to find the yz plane I set x to 0? the thing is then all intersection type will be a circle.. and I am not sure if that's the right answer

 

[Dick]

The radius of the sphere is sqrt(20). The center of the sphere is (4,3,-2). The farthest coordinate plane is the yz plane. The center of the sphere is 4 units away from that. sqrt(20)>4. They are all circles.

 

[-EquinoX-]

ok, so what if the question is

"How does the sphere intersect each of the x-axis, y-axis, z-axis"?

 

[-EquinoX-]

all of them? how do I find the equation though?

 

[Dick]

all of them? how do I find the equation though?

I retracted that quick (and wrong) answer. For the x-axis you have to figure out if the circle in the x-y plane intersects the x-axis. More simply, where you hit the x-axis is where y=0, and z=0. Etc.

 

[-EquinoX-]

if that's the case then what's left is just (x-4)^2 = 20, correct? in that case it will intersect at two points as it's a quadratic equation

 

[Dick]

No, what's left if y=0 and z=0 is (x-4)^2+3^2+2^2=20.

 

[-EquinoX-]

yes that's true, either way all will be in that form and solving it is just basically a quadratic equation and we get two points for each plane (xy-plane, yz-plane, xz-plane), correct?

 

[Dick]

No. That's why I retracted my answer. The intersections with the z-axis have to satisfy (z+2)^2+3^2+4^2=20. Now what?

 

[-EquinoX-]

z^2+4z+9 to make it more simple, right... and that is a quadratic equation, so why can't we get two points out of it?

 

[Dick]

(z+2)^2=(-5) to put it even more simply. Some quadratic equation don't have any real solutions. Face it.

 

[-EquinoX-]

oh, you're right.. we can't take the square root of 5... so for a special case for the z-axis, it doesn't intersect anywhere, not even at a point.. right?

 

[Dick]

Right. You meant sqrt(-5), right?

 

[-EquinoX-]

the equation with the intersection at the y-axis is simply:
= (y-3)^2 = 0
= y-3 = 0
= y = 3

so it just intersects at a point

 

[Dick]

the equation with the intersection at the y-axis is simply:
= (y-3)^2 = 0
= y-3 = 0
= y = 3

so it just intersects at a point

Sure.