A circle of radius 4 is contained in the plane y=7.

Both circles lie on the surface of a sphere.

What is the radius of the sphere?

For one second, I thought these were two DISKS that were TANGENT to a sphere, so I was thinking "duh! The answer is (7-3)/2 = 2" but then I realized my silliness =) Did anyone else do that? Just curious.

By the way, how is this different from the Math on the Playground problem? I guess it just takes your brain another step to realize you can treat this 3D problem like a 2D problem, but it’s essentially the same, yes?

Anyways, in these problems I have always been troubled by the fact that I don’t know if the two circles (or roadways) are in the same hemisphere (or halfcircle). Let me show you what I mean:

r = radius of smaller circle = 2
R = radius of larger circle = 4
H = perpendicular distance between two circles = 7-3 = 4
a = perpendicular distance between center of sphere and smaller circle (r)
b = perpendicular distance between center of sphere and larger circle (R)
S = radius of sphere

If circles r and R are in the same hemisphere, then H = a-b. But if the circles are in different hemisphere, then H = a+b.

I solved for both situations, and it turns out that, even though the equations are technically different, it doesn’t matter which you guess due to squaring certain terms. Ready? Here goes:

Always true:
S^2 = a^2 + r^2
S^2 = b^2 + R^2

S^2 = a^2 + r^2 = b^2 + R^2
a^2 – b^2 = R^2 – r^2

First, assume that the two circles are in DIFFERENT hemispheres. So H = a+b
So a = H-b
(H-b)^2 – b^2 = R^2 – r^2
H^2 – 2Hb + b^2 – b^2 = R^2 – r^2
H^2 – R^2 + r^2 = 2Hb
b = (H^2 – R^2 + r^2)/(2H)
So S^2 = b^2 + R^2 = (H^2 – R^2 + r^2)^2/(2H)^2 + R^2

Now, assume that the two circles are in the SAME hemisphere. So H = a-b
So a = H+b
(H+b)^2 – b^2 = R^2 – r^2
H^2 + 2Hb + b^2 – b^2 = R^2 – r^2
H^2 – R^2 + r^2 = -2Hb
b = (H^2 – R^2 + r^2)/(-2H)
So S^2 = b^2 + R^2 = (H^2 – R^2 + r^2)^2/(-2H)^2 + R^2

But since (-2H)^2 = (2H)^2, we’ll get the same answer no matter which way we guessed it.

Anyways, S^2 = (4^2 – 4^2 + 2^2)^2/(2*4)^2 + 4^2 = (4)^2/(8)^2 + 16 = 1/4 + 16 = 16.25.

So the radius is sqrt(16.25)
a = sqrt(12.25) = sqrt(49/4) = 7/2 = 3.5
b = sqrt(0.25) = sqrt(1/4) = 1/2 = 0.5

So it turns out that the two circles are in different hemispheres, just FYI.