in the give figure move the line that joins the centres of the circles downwards(or upwards).
now you will get a right angle triangle whose one side is x+y, other side is x-y and the third side can be found (pythagoras theorem)which is the length of segment of rubber band.by symmetry other other segment of band will also have same length.
nikhil's formula will work in rough cut situations.
However, if you need an exact fit, try this:
Let R be the radius of the larger circle.
Let r be the radius of the smaller circle.
Extend the two tangent lines until they intersect.
Let theta be the angle at the intersection.
LengthOfRubberBand=
Theta in radians.
Nikhil must explain how he got the sum of the lengths of the two arcs as
pi (x + y). The explanation about the part of the rubber band between the two circles is OK.
Aidan, I think you are introducing a new variable "theta". Your answer must be only in terms of x and y.
Vipin
Sorry, I never saw that condition until now.
However, it is easily solved:
Let x be the radius of the larger circle.
Let y be the radius of the smaller circle.
Thus:
LengthOfRubberBand = x[pi + (1/1)(x^1/1) + (1/2)(x^3/3) + (1/2)(3/4) x^5/5) + (1/2)(3/4)(5/6)(x^7/7) + (1/2)(3/4)(5/6)(7/8)(x^9/9) + (1/2)(3/4)(5/6)(7/8)(9/10)(x^11/11) + (1/2)(3/4)(5/6)(7/8)(9/10)(11/12)(x^13/13)] + y(pi - [ (1/1)(((x-y)(x+y))^1/1) + (1/2)(((x-y)(x+y))^3/3) + (1/2)(3/4)(((x-y)(x+y))^5/5) + (1/2)(3/4)(5/6)(((x-y)(x+y))^7/7) + (1/2)(3/4)(5/6)(7/8)(((x-y)(x+y))^9/9) + (1/2)(3/4)(5/6)(7/8)(9/10)(((x-y)(x+y))^11/11) + (1/2)(3/4)(5/6)(7/8)(9/10)(11/12)(((x-y)(x+y))^13/13)] ) + 4 sqrt(xy)
There all in terms of x and y.
ok?