1. Find the ellipse of smallest area which circumscribes 2 non-overlapping unit circles.
2. Find the ellipse of smallest area which circumscribes 3 non-overlapping unit circles.
let the circles be on the x-axis equidistant (d) from the origin
 (x+/-d)^2 + y^2 = 1
The ellipse is then in standard form
 x^2/a^2 + y^2/b^2 = 1
Solve  for y and sub. into , simplify and set equal to zero
 (b^2-a^2)x^2 + (2a^2d)x + (a^2 -a^2d^2-a^2b^2) = 0
This will have a single solution for x when the discriminant = 0
 (2a^2d)^2 - 4(b^2a^2)(a^2-a^2d^2-a^2b^2) = 0
Solve for d^2
 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 . To minimize the area, we must minimize a*b. Solve this for a
 a*b = b*sqrt(((d^2-1)b^2+b^4)/(b^2-1))
The problem is then to minimize this expression. This would require finding the derivative of  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?
Posted by Jer
on 2004-11-29 19:19:37