I think your problem reduces to finding a circel through
three points <-r1,0> <r1,0> and <0,h1>
The standard way to solve that is taking two lines
through the middle of the lines and perpendicular to those
lines. In your case it is even more simple, just solve
where one of the lines cuts the y-axis:

<-r1/2,h1/2>+ c*<-h1,r1>=<0,h2>

this will give

c=1/2*r1/h1

and therefore

h2=-1/2*(h1-r1*r1/h1)

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)
>

Post a reply to this message