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

[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^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

[4]     (2a^2d)^2 - 4(b^2a^2)(a^2-a^2d^2-a^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^2-1)b^2+b^4)/(b^2-1))

from which

[7]     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 [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 2004-11-29 19:19:37