another way to solve this:
for the 2nd circle with an origin an <0,-h2> to go through
<r1,0> ir must have a radius r2 of sqrt(r1^2+h2^2) and also
because it must go through <0,h1> of h1+h2
solving
r2^2 + h2^2 = (h1+h2)^2
will give you the answer.

andrel

Tom Melly wrote:

> See: http://www.uk.ccl.com/tom/bi_circ.gif for accompanying illustration.
> 
> Given a circle of radius r1 with a centre at <0,0,0>, how do I calculate the
> radius (r2) and vertical displacement (h2) of a second circle, such that the two
> circles bisect each other at y=0 with a particular height (h1) of the second
> circle at its apex?
> 
> Hopefully the attached link will clear up any ambiguities in what I'm asking.
> 
> (I have to admit that the question has become slightly academic, since I've
> accomplished roughly what I needed with two identical circles, and some scaling
> of y on the second circle)
>

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