Geodesics of hyperbolic plane

Question

I just found (or: I think that I found) the geodesics of the upper, closed half plane of $\mathbb R^2$. To verify my solution: Is it correct that the geodesics are the circles and lines which meet the unit circle at right angles?

Thanks a lot for the confirmation (or the correction, if I'm wrong...)

Edit: Here is what I have so far (this should be correct, isn't it?):

Answer 1

It is not correct that the geodesics of the upper half plane $\mathbb{H}=\{(x,y)\in\mathbb{R}^2:\;y>0\}$ with metric $g$ given by $g_{xx}=g_{yy}=1/y^2$ and $g_{xy}=g_{yx}=0$ the circles and lines which meet the unit circle at right angles. Instead, they are the circles and lines that meet the $x$-axis (the line $y=0$ in $\mathbb{R}^2$) at right angles.

A curve $\gamma:I\subset\mathbb{R}\rightarrow \mathbb{H}:s\mapsto(x_1(s),x_2(s))$ is a geodesic if it satisfies the geodesic equation: \begin{equation} \frac{d^2x_k}{ds^2} + \sum_{i,j=1}^2\Gamma^k_{ij}\frac{dx_i}{ds}\frac{dx_j}{ds}=0, \end{equation} for $k=1,2$ and where $\Gamma^k_{ij}$ are the Christoffel symbols, which in this case are \begin{equation} \Gamma^1_{12}=\Gamma^1_{21}=\Gamma^2_{22}=-1/y;\;\; \Gamma^2_{11}=1/y;\;\;\; \Gamma^1_{11}=\Gamma^2_{12}=\Gamma^2_{21}=\Gamma^1_{22}=0. \end{equation} The geodesic equations are then the following system of differential equations (writing $x=x_1$ and $y=x_2$): \begin{equation} \frac{d^2x}{ds^2}-2\frac{1}{y}\frac{dx}{ds}\frac{dy}{ds}=0;\;\;\; \frac{d^2y}{ds^2}+\frac{1}{y} \left[\left(\frac{dx}{ds}\right)^2-\left(\frac{dy}{ds}\right)^2\right]=0 \end{equation} For $\frac{dx}{ds}\neq 0$ the solutions satisfy: $x^2+y^2-ax=b$, where $a$ and $b$ are constants. Those are circles with center in the line $y=0$, and thus meeting that line at right angles. The solutions for $\frac{dx}{ds}=0$ are just vertical lines.

Answer 2

It is possible to write down the solution in full parametric form. It is parametrized with the intial conditions: $x(0)=x_0$, $y(0)=y_0$, $x'(0)=u_0$, $y'(0)=v_0$ and additionally the constraint $u_0^2+v_0^2=y_0^2$ is implied.

For the circle:

$x(\lambda) = x_0+\frac{u_0 y_0\sinh\lambda}{y_0\cosh\lambda-v_0\sinh\lambda}$

$y(\lambda)=\frac{y_0^2}{|y_0\cosh\lambda-v_0\sinh\lambda|}$

For $u_0=0$ it follows that $v_0=y_0$ and the straight line is recovered from the general solution:

$x(\lambda)=x_0$

$y(\lambda)=y_0 \exp(\lambda)$