Math Help -> Puzzles -> Two circles and a sphere -> Answer

Two Circles and a Sphere

Two parallel planes are 4 units apart. One plane contains circle A, with radius 2. The other plane contains circle B, with radius 4. If both circles lie on the surface of a sphere, what is the radius of the sphere?

Answer

Here is a diagram showing a "side view" of the circles and sphere, sliced CHOP! by a plane perpendicular to the two given planes (parallel to your computer monitor) going through the center of both circles and the sphere. (These centers must be collinear given the symmetry of the figures.) We are left with circle "O" and two parallel chords (AB and CD), 4 units apart. Chord AB has length 4, and chord CD has length 8.

By connecting the "tops" and "bottoms" of the chords, we make a quadrilateral ABCD -- but not just any quadrilateral; this is a "cyclic quadrilateral" ( which just means it can be inscribed in a circle). Ptolemy's theorem says that the product of the diagonals equals the sum of the product of opposite sides.

The diagonals are each sqrt(6²+4²) = sqrt(52) in length, so their product is 52.

The two given chords AB and CD are 4 and 8 in length, so their product is 32.

That means the product of the two chords we constructed (BC and DA) is 52-32=20. Since BC=DA, the length of each is sqrt(20).

The circumradius of a cyclic quadrilateral with consecutive sides a, b, c, d is given by

R = (1/4)sqrt((ac+bd)(ad+bc)(ab+cd)/((s-a)(s-b)(s-c)(s-d)))

In this case, let a=4, b=sqrt(20), c=8, and d=sqrt(20). Then it's just a matter of simplifying the formula:

R=(1/4)sqrt((32+20) (4sqrt(20)+8sqrt(20)) (4sqrt(20)+8sqrt(20)) / ((6+sqrt(20)-4) (6) (6+sqrt(20)-8) (6)))
R=(1/4)sqrt((52) (12sqrt(20)) (12sqrt(20)) / (36(sqrt(20)+2) (sqrt(20)-2)))
R=(1/4)sqrt(52*12*12*20 / (36*16))
R=(1/4)sqrt(52*20 / 4)
R=(1/4)sqrt(52*5)
R=(1/2)sqrt(65)

Another Way

Another way to solve this problem is to put it on an x-y grid. Let points A, B, C, and D be

(0,-2), (0,2), (4,4), and (4,-4).

Now, pick any three points on the circle, for example: (0,-2), (0,2), and (4,4). These three points determine a circle whose center is on the x-axis. When the three points are (a,b), (c,d), and (e,f) the x-value of the center, h, is given by the formula

h = ((a²+b²)(f-d) + (c²+d²)(b-f) + (e²+f²)(d-b)) / (2(a(f-d)+c(b-f)+e(d-b)))

Plugging in the numbers, and solving for h gives

h = ((0+4)*(4-2) + (0+4)*(-2-4) + (16+16)*(2+2)) / (2*(0*(4-2)+0*(-2-4)+4*(2+2)))
h = ((4)*(2) + (4)*(-6) + (32)*(4)) / (2*(0+0+4*(4)))
h = (8 - 24 + 128) / (2*4*4)
h = 112 / 32
h = ⁷/₂

Now the radius is the distance from (⁷/₂,0) to any of the four points A, B, C, or D. I'll pick (0,2). The distance is sqrt(49/4+4) = sqrt(65/4), which is the same answer as above.

Internet references

http://mathworld.wolfram.com/CyclicQuadrilateral.html

Related pages in this website

Ptolemy's theorem
Finding a circle given three points

The webmaster and author of the Math Help site is Graeme McRae.
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