How to identify all the right circular cones passing through six arbitrary points I have this interesting question.  Given $6$ arbitrary points, I want to identify all the possible circular cones passing through them.
The equation of a right circular cone whose vertex is at $\mathbf{r_0}$ and whose axis is along the unit vector $\mathbf{a}$, and whose semi-vertical angle is $\theta$ is given by
$$ (\mathbf{r} - \mathbf{r_0})^T ( (\cos \theta)^2 \ \mathbf{I} - \mathbf{aa}^T ) (\mathbf{r} - \mathbf{r_0}) = 0 $$
where $\mathbf{r}$ is the position vector of a point on the surface of the cone.  Counting the number of parameters, we have $3$ for $\mathbf{r_0}$, $2$ for $\mathbf{a}$ and $1$ for $\theta$.  We therefore need at least $6$ points on the surface of the cone to specify it.
Question:  What is the procedure for extracting the parameters of a right circular cone passing through $6$ arbitrary points?
What I have tried:  I have parameterized the axis unit vector as
$ \mathbf{a} = ( \sin t \cos s, \sin t \sin s , \cos t ) $
and written
$\mathbf{r_0} = (x, y, z) $
Now define the functions
$ f_1 =  (\mathbf{r_1} - \mathbf{r_0})^T ( (\cos \theta)^2 \ \mathbf{I} - \mathbf{aa}^T ) (\mathbf{r_1} - \mathbf{r_0})  $
$ f_2 =  (\mathbf{r_2} - \mathbf{r_0})^T ( (\cos \theta)^2 \ \mathbf{I} - \mathbf{aa}^T ) (\mathbf{r_2} - \mathbf{r_0})  $
$ f_3 =  (\mathbf{r_3} - \mathbf{r_0})^T ( (\cos \theta)^2 \ \mathbf{I} - \mathbf{aa}^T ) (\mathbf{r_3} - \mathbf{r_0})  $
$ f_4 =  (\mathbf{r_4} - \mathbf{r_0})^T ( (\cos \theta)^2 \ \mathbf{I} - \mathbf{aa}^T ) (\mathbf{r_4} - \mathbf{r_0})  $
$ f_5 =  (\mathbf{r_5} - \mathbf{r_0})^T ( (\cos \theta)^2 \ \mathbf{I} - \mathbf{aa}^T ) (\mathbf{r_5} - \mathbf{r_0})  $
$ f_6 =  (\mathbf{r_6} - \mathbf{r_0})^T ( (\cos \theta)^2 \ \mathbf{I} - \mathbf{aa}^T ) (\mathbf{r_6} - \mathbf{r_0})  $
Now using the multivariate Newton-Raphson method I could iterate to find a solution for the parameter vector $(t, s, x, y, z , \theta )$ that will make $f_1, f_2, f_3, f_4, f_5, f_6$ all zero.
This works, but it's iterative and at best converges to one of the possible right circular cones.
Is there a way to generate all the possible right circular cones that are solutions, i.e. passing the $6$ given points?

Notation. Let vector $c$ be the center of cone and unit vector $a$ be its axis of symmetry. Let the angular aperture (the angle each generator makes with the axis) be $\theta$.
First let's derive a system of linear equations that $c$ satisfies.
Any known point $x$ on the cone satisfies
$|x-c|^2= (A\cdot(x-c))^2$ where vector $A= \sec \theta  \ a$ has known length but unknown direction.
Likewise any second known point $x'$ satisfies the same type of equation.
Subtracting one equation from the other, obtain
(1) $|x|^2 - |x'|^2 -  2c\cdot( x-x') = [A\cdot (x-c)]^2-[A\cdot (x'-c)]^2$
and the RHS factors as
$ [A\cdot(x-c) -A\cdot (x'-c)][ A\cdot(x-c) + A\cdot(x'-c)]$
$= [A\cdot (x-x')][-2 A\cdot c + A\cdot(x+x')]$
As functions of $c$, both the left and right sides of (1) are linear (actually affine) in $c$. (And as function of $A$, (i) is a quadratic in the components of $A$). We can obtain three such linear  inhomogeneous equations by cyclically permuting any three known points $x,x',x''$ on the conic. Inverting that linear system by e.g. Cramer's Rule, we solve for $c$,still in terms of the unknown vector $A$. Thus each entry of $c$ is a rational function of the entries of $A$. (In fact the denominator is the determinant of the linear system whose coefficients are quadratics in $A$. Thus the determinant has degree 6 in $A$. The numerator has degree 4 in $A$.)
Once $c$ is so determined as an explicit function of $A$, one can obtain another set of equations that $A$ must satisfy by selecting another triplet $(X,X', X'')$ of known points on the conic, and writing down  equations (i) for that new triplet. At this stage we have eliminated $c$ and therefore have three rational equations in $A$ to be solved, subject of course to the condition that $|A|=\sec\theta$.
At this final stage, perhaps you would find it fruitful to parametrize that sphere of radius $|A|$ using stereographic projection, so that $A$ becomes a rational function of two rectangular coordinates.  This reduces the problem to algebraic equations in two unknowns.  That is, you seek the intersection points of some algebraic curves.
Good luck!
