# Plane-Sphere Intersections, how do they work? I've just done a few excercices regarding this "subject", and I'm near desperation, first of all, according to my book, a section of a plan defined...

Plane-Sphere Intersections, how do they work?

I've just done a few excercices regarding this "subject", and I'm near desperation, first of all, according to my book, a section of a plan defined with "Y = 3" intersecting a sphere defined by "(X - 1)^2 + Y^2 + (z - 3)^ 2 <= 9" has the coordinates of (1, 3, 3), what? How would I find this out? I have no idea. Another thing the book tells me is that the condition of a sphere with the center of (0, 2, 0) tangent to the plan Z = 4 is "x^2 + (Y -2)^2 + Z^2 <= 16", once again, what? How would I find this out? I have no idea. I feel like I'm missing a huge gap in the logic behind these excercices, I'm just baffled when I look at the solutions of these excercices. Can someone help me? I just need to find out the logic so I can do excercices like these myself.

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### 1 Answer

The equation (x-g)^2+(y-h)^2 +(z-k)^2 < = r^2 all the set of points on and inside a sphere with centre (g,h,k) and radius r.

So (x-1)^2+y^2+(z-3)^2 < = 9 = 3^2 represents a sphere with centre at (1, 0, 3) and radius 3.

The section of a plane y = 3 with the sphere (x-1)^2+y^2+(z-3)^2 is a circle with centre (x,3,y). So (x-3)^2+3^2+(z-3)^2 = 0 implies (x-1)^2 + 3^2 +(z-3)^2 < = 3. (x-1)^2+(z-3)^2 = 0. This implies that (1, 3, 3) is the centre of the circle with zero radius. Or the plane y = 3 touches the sphere with centre at (1,3,3).

Now we consider the equation of a sphere with centre is (0,2,0). The sphere has the equation (x-0)^2+(y-2)^2+(z-0)^2 = r^2, or x^2+(y-2)^2+z^2 <= r^2...(1) where r is the radius of the sphere.

Now if z = 4, is a tangent plane to the sphere, with centre (0,2,0) then the distance from the tangent plane to the centre of the sphere must be radius. The point on the surface of the plane z = 4 touching the sphere with centre (0,2,0) is (0, 2, 4).

So the distance between the (0,2,0) and (0,2,4) is given by r^2 = (0-0)^2+(2-2)^2+(0-4) ^2 = 4^2. So r = 4. We substitute this value of r = 4 in the equation of the sphere (1) and get:

x^2+(y-2)^2 +z^2 < = 4^2 = 16.