Using this technique I came up with HUGELY complicated equations for one (???) point.
I haven't put this into pov yet, but I fear that my the values of the variables in my
equations will boild down to:

(0-0)/(0-0)
.

Ken Matassa wrote:

> Christoph Hormann wrote:
> >
> > "Greg M. Johnson" wrote:
> > >
> > > As I think about it the five possible cases are:
> > >     no intersection
> > >     1 pt
> > >     2 pts
> > >     a circle
> > >     a sphere
> > >
> > > For my application at hand, I am interested in the 2 pt scenario, if that
> > > makes the answer easier for anyone to find, thanks.
> > >
> >
> > I'm not so sure, it has been some time, since i did that kind of math stuff, but
> > the general equation of a sphere is:
> >
> >  ( x - xm )^2 + ( y - ym )^2 + ( z - zm )^2 - r^2 = 0
> >
> > with <xm, ym, zm> the center of the sphere and r the radius.
> >
> > Make it 3 different spheres and all equations being zero -> you should get the
> > result.
> >
> > (I'm not totally sure about this, so feel free to correct me if there is a
> > mistake)
> >
> > BTW, i suspect this does not solve your problem, because it's not a numerical
> > method, but maybe it helps somehow.
> >
> > Christoph
> >
> > --
> > Christoph Hormann <chr### [at] gmxde>
> > Homepage: http://www.schunter.etc.tu-bs.de/~chris/
>
> You're on the right track.
>
> You need three equations in three unknowns: ( In VERY general terms!)
>
> a(x1)^2 + b(y1)^2 + c(z1)^2 = 0
> d(x2)^2 + e(y2)^2 + f(z2)^2 = 0
> g(x3)^2 + h(y3)^2 + i(z3)^2 = 0
>
> You now simultainously solve the three equations. If they are liniarly
> independent the there will be 2 uneque salutions, the points of
> intersection. Now there is a gotch in this. I know how to do this with
> liniar equations, BUT, I don't know if it can be donw like this with
> quadratic equations. Probubly have to look in an advanced math textbook.
> I don'e recall second or higher order equations being covered in my
> liniar algebera course.
>
> Ken Matassa

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