Line inside the hyperboloid going through a point of a circle I am having trouble with an absurdly simple problem. It has been a long time since I last dealt with this kind of problem.
Consider the one-sheet hyperboloid given by 
$$
                          x^2+y^2-a^2z^2=c^2
$$
and let $(X,Y,0)$ be a point of the circle with radius $c$.
I want to find the vector $v=(A,B,C)$ such that the locus $\{(X,Y,0)+vt:t\in\mathbb{R}\}$ is a line contained in the hyperboloid.
The equations I got are
$$
           \begin{cases}
x^{2}+y^{2}-a^{2}z^{2}=c^{2}\\
x=At+X\\
y=Bt+Y\\
z=Ct+0
\end{cases}
$$
I tried to mess with them a bit but couldn't solve my problem. Worse: I do not know how I can deal with the $t$ (I do not want $A,B,C$ to depend on it, of course).
 A: You're nearly there. Substituting your expressions for $x, y, z$ into the equation for the hyperboloid gives an polynomial in $A, B, C, t$.
$$(A t + X)^2 + (B t + Y)^2 - a^2 (C t)^2 = c^2.$$
Since all points on the line must be contained in the hyperboloid, this equation must hold for all times $t$, we can collect and compare like terms in $t$. Rearranging gives
$$(A^2 + B^2 - a^2 C^2) t^2 + 2(A X + B Y) t + (X^2 + Y^2) = c^2.$$

Comparing like terms in $t$ gives $$\left\{ \begin{array}{rcl}A^2 + B^2 - a^2 C^2 &=& 0 \\ 2 (A X + B Y)       &=& 0 \\ X^2 + Y^2           &=& c^2 \\ \end{array} . \right.$$ The third equation tells us something we already know, namely that $(X, Y, 0)$ sits on the given circle of radius $c$. This leaves two equations in three unknowns ($A, B, C$), so generically one expects there to be one degree of freedom in the solution. We could have expected something like this anyway, since if $v = (A, B, C)$ is a solution, so is any nonzero multiple $\lambda v$, as they both determine the same line.

A: One of the easier results to establish by differential geometry for a hyperboloid of one sheet with geodesics (which are also asymptotic) are, as a function of arc length s  on surface with $s = v\,t$ and 
Clairaut radius =$r_o = r\, \sin \psi,$ start (mid neck) radius at zero slope  $ \phi=0,\, r=c ,$ the start angle between geodesic and meridian $\psi=\alpha$. We have the constant
$$ \cos \alpha = \cos \phi\, \cos \psi, \tag1$$ 
and in polar coordinates we simply have
$$ r = \sqrt{c^2 + (s r_0 /c)^2 },\, \theta= \cos^{-1} \frac{c}{r}, z= s\cdot \cos \alpha \tag2 $$
