On Mon, 2 Feb 2004 12:11:37 -0000, Tom Melly <tom### [at] tomandlucouk> 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)
>

<code>
//All the Math

#declare x1=0;
#declare y1=-r1;
#declare x2=-h1;
#declare y2=0;
#declare x3=0;
#declare y3=-r1;

#declare ma=(y2-y1)/(x2-x1);
#declare mb=(y3-y2)/(x3-x2);

#declare myx=((ma*mb*(y1-y3))+(mb*(x1+x2))-(ma*(x2+x3)))/(2*(mb-ma));

#declare NewRadius=(h1)+myx;

</code>

Is the way I found to do it; finds the centre of the new circle from three 
points. You might want to check my math though:)

--
Phil

-- 
All thoughts and comments are my own unless otherwise stated and I am 
happy to be proven wrong.

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