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).

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
$$ r = \sqrt{c^2 + (s r_0 /c)^2 },\, \theta= \cos^{-1} \frac{c}{r}, z= s\cdot \cos \alpha \tag2 $$
