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Re: Overlapping circles
Posted:
Jun 6, 2001 6:03 PM


W.H. wrote:
> Two circles of diameter (d) and (2*d) overlap (not over the > centres) so that they have a common chord of length (c). The > measurement across the two circles perpendicular to the chord > is (s) distance. I need to find the diameters in terms of (c) > and (s).
Let's work with arbitrary diameters d1 and d2 rather than d and 2d. Later we will substitute d1 = d and d2 = 2d.
You assume the circles intersect. This means that s < d1 + d2.
You also assume that one circle is not completely inside the other. This means s > d1 and s > d2, which implies that 2s > d1 + d2. Therefore s > d1/2 + d2/2.
Let O1 and O2 be the centers of the two circles. Let E be one of their intersection points. Focus on the triangle O1EO2.
It's easy to see that the distance between O1 and O2 is s  d1/2  d2/2.
Therefore the three sides of the triangle are d1/2, d2/2 and s  d1/2  d2/2. We can express its area by plugging these into Heron's formula. We get:
(1) Area^2 = s (s  d1) (s  d2) (d1 + d2  s) / 16.
Note that the right hand side is positive due to the inequalities stated above.
At the same time, the area can be expressed as one half of the base times height. But the base is s  d1/2  d2/2 and the height is c/2. Therefore
Area = (1/2) (c/2) (s  d1/2  d2/2)
which gives:
(2) Area^2 = [ (1/2) (c/2) (s  d1/2  d2/2) ]^2.
Now equate the two different expressions for Area^2 in equations (1) and (2):
(3) s (s  d1) (s  d2) (d1 + d2  s) / 16 = [ (1/2) (c/2) (s  d1/2  d2/2) ]^2
Equation (3) establishes a relationship between the four quantities d1, d2, s, and c. Given any three of them, you can calculate the remaining one.
This completes the analysis of the general case. You have asked specifically for d1 = d and d2 = 2d. In that case, equation (3) reduces to:
 4 s (s  d) (s  2d) (s  3d) = c^2 (2s  3d)^2 .
As you see, this is a cubic equation in d. Apply any of the known methods to solve it.
 Rouben Rostamian <rostamian@umbc.edu>



