Geodesics on a pseudosphere I am trying to show the following:
Let
\begin{equation}
\gamma(t) = \begin{pmatrix}
\frac{1}{t} \\
\sqrt{1 - \frac{1}{t^2}} + \cosh^{-1}(t)
\end{pmatrix}
\end{equation}
for $t \geq 1$ be the unit-speed parametrization of the tractrix and
\begin{equation}
\sigma(u,v) = \begin{pmatrix}
\frac{1}{u} \cos v \\
\frac{1}{u} \sin v \\
\sqrt{1 - \frac{1}{u^2}} + \cosh^{-1}(u)
\end{pmatrix}
\end{equation}
the surface of revolution obtained by rotating the tractrix around the z-axis. Now I want to show that the geodesics on the pseudosphere (tractroid in this case) are represented in the $(u,v)$-coordinates by arcs of circles centered on the $v$-axis.
The first thing that I noted is that the curve is not unit-speed parametrized.
So I calculated the coefficients of the first fundamental form (for a surface patches)
\begin{align*}
  E(u,v) &= \dot{f}^2(u) + \dot{g}^2(u) = \frac{3 + u^2}{u^2(u^2 - 1)}, \\
  F(u,v) &= \langle \sigma_u(u,v), \sigma_v(u,v) \rangle = 0, \\
  G(u,v) &= \frac{1}{u^2} = f^2(u)
\end{align*}
and now the geodesic equations are
\begin{align*}
\frac{d}{dt}\left( E \dot{u} + F \dot{v} \right)  &= \frac{1}{2}\left( E_u \dot{u}^2 + 2 F_u \dot{u} \dot{v} + G_u \dot{v}^2 \right), \\
\frac{d}{dt}\left( F \dot{u} + G \dot{v} \right) &= \frac{1}{2}\left( E_v \dot{u}^2 + 2 F_v \dot{u} \dot{v} + G_v \dot{v}^2 \right)
\end{align*}
therefore
\begin{align*}
 \frac{d}{dt} \left(\frac{3 + u^2}{u^2(u^2 - 1)}  \dot{u} \right) &= -\frac{4 u \dot{u}^2 }{(u^2 - 1)^2} - \frac{\dot{v}^2 - 3 \dot{u}^2}{u^3}, \\
\frac{d}{dt}\left( \frac{1}{u^2}\dot{v} \right) &= 0
\end{align*}
solving the second gives
\begin{equation*}
  \dot{v} = C u^2
\end{equation*}
which can also be obtained by clairauts theorem
where $\psi$ is the angle between $\dot{\gamma}$ and the meridians of the surface
\begin{equation*}
  \dot{v} = \frac{\sin \psi}{f} = u \sin \psi
\end{equation*}
and $f \sin \psi$ being constant gives
\begin{equation*}
  f^2 \dot{v} = \frac{1}{u^2} \dot{v} = f \sin \psi \stackrel{!}{=} \mbox{const}.
\end{equation*}
Plugging this $\dot{v}$ into the first geodesic equation leaves us with
\begin{equation*}
  \frac{d}{dt} \left(\frac{3 + u^2}{u^2(u^2 - 1)}  \dot{u} \right) = \frac{\dot{u}\left(3 - \frac{4 u^4}{(u^2 - 1)^2} \right)}{u^3} - C^2 u.
\end{equation*}
From this I can't figure out how the geodesics are arcs of circles centered on the $v$-axis. So I tried using the fact that
\begin{equation}
I(\dot{\gamma},\dot{\gamma}) \stackrel{!}{=} 1
\end{equation}
which gives
\begin{equation}
E \dot{u}^2 +  G \dot{v}^2 = 1
\end{equation}
or
\begin{equation}
\dot{u}^2 \frac{3 + u^2}{u^2 - 1} +  C^2 u^4 = u^2
\end{equation}
thus
\begin{equation}
\dot{u} = \pm \frac{\sqrt{u^2(u^2C^2 - 1)}}{\sqrt{ \frac{3 + u^2}{1 - u^2} }}.
\end{equation}
From this I would get an expression for $\frac{\dot{v}}{\dot{u}}$ hence by separation of variables $(v - v_0) = (\ldots)$ where the right hand side involves an elliptic integral.
I would expect that from $\frac{\dot{v}}{\dot{u}} = F(u)$ where $F$ is some function depending only on $u$, I can use separation of variables to get something like
\begin{equation}
(v-v_0)^2 + u^2 = \frac{1}{C}
\end{equation}
but from the geodesic equation and the ODE it is not possible to obtain such a form. If I would switch the sign from $\cosh^{-1}(t)$ to $-\cosh^{-1}(t)$ this would work.
I would be thankful if anybody can give me a hint.

Comment
If $ (f, v) $ are polar coordinates it is in hyperbolic geodesic representation that have circles passing through origin and centered on the x-axis... or any radial line but not the lines on 2-D surface in 3-space.

