1. Find the ellipse of smallest area which circumscribes 2 nonoverlapping unit circles.
2. Find the ellipse of smallest area which circumscribes 3 nonoverlapping unit circles.
let the circles be on the xaxis equidistant (d) from the origin
[1] (x+/d)^2 + y^2 = 1
The ellipse is then in standard form
[2] x^2/a^2 + y^2/b^2 = 1
Solve [1] for y and sub. into [2], simplify and set equal to zero
[3] (b^2a^2)x^2 + (2a^2d)x + (a^2 a^2d^2a^2b^2) = 0
This will have a single solution for x when the discriminant = 0
[4] (2a^2d)^2  4(b^2a^2)(a^2a^2d^2a^2b^2) = 0
Solve for d^2
[5] d^2 = a^2  b^2 + 1  a^2/b^2
These are the values of a and b for any ellipse which will circumscribe the circles from [1]. To minimize the area, we must minimize a*b. Solve this for a
[6] a=sqrt(((d^21)b^2+b^4)/(b^21))
from which
[7] a*b = b*sqrt(((d^21)b^2+b^4)/(b^21))
The problem is then to minimize this expression. This would require finding the derivative of [7] and setting it equal to 0 and solving the resulting equation. I just couldn't handle the algebra from this point.
For specific values of d it is not too hard to approximate the ellipse of minimum area.
d=2, a=3.3465, b=1.316 area=13.836
THREE CIRCLES!!!??? Forget it! How would you know where to put them to guarantee there even IS a circumscribing ellipse?
Jer

Posted by Jer
on 20041129 19:19:37 