 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Ellipses and Circles (Posted on 2004-11-29) 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.

 No Solution Yet Submitted by Brian Smith Rating: 4.0000 (4 votes) Comments: ( Back to comment list | You must be logged in to post comments.) As far as I plan to get | Comment 2 of 6 | 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=sqrt(((d^2-1)b^2+b^4)/(b^2-1))

from which

     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?

-Jer

 Posted by Jer on 2004-11-29 19:19:37 Please log in:

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