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,...

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.