There is a general fact in Riemannian geometry which follows from "standard" existence and uniqueness theorems in the study of ODEs. Mainly, given any point $p$ in a Riemannian manifold $M$ and given any $v\in T_pM$, there is a unique geodesic defined for some open neighborhood of time around 0 starting at $p$ in the direction of $v$.
Assuming this, it's not too hard to show that you've found all the geodesics. You just need to verify that for any point $p$ in the hyperbolic plane and tangent vector, either the vector points up/down (so a vertical line geodesic has $v$ as tangent vector), or some semicircle through $p$ hitting the $x$-axis perpendicularly does.
Finally, to show there is a unique geodesic between any two points, one argues as follows:
Given $p$ and $q$, if $q$ and $p$ have the same $x$ coordinate, a vertical line passes through them. Further, no semicircle hitting the $x$-axis perpendicularly can go through both $p$ and $q$ since the semicircle is the graph of a function, so passes the vertical line test.
Next, given $p$ and $q$ with different $x$ coordinates, it's clear that a vertical line can't connect them, so we need only find all semicircles passing through both perpendicular to the $x$-axis. Algebraically, proceed by noting the equation of such a circle is $(x-a)^2 + y^2 = r^2$ for some $a$ and $r$. Plugging in $(x,y) = (p_1,p_2)$ and $(x,y) = (q_1,q_2)$ respectively and setting the two equal to each other (since they both equal $r^2$), one gets $$(p_1-a)^2 + p_2^2 = (q_1-a)^2 + q_2^2.$$
By simplifying this, one sees the $a^2$ cancels out and so one gets an equation linear in $a$, so $a$ is uniquely determined. (The assumption that $p_1\neq q_1$ means that the equation has the linear piece). Knowing $a$ easily implies that $r$ is uniquely determined (keeping in mind that $r \geq 0$). This shows that there is a unique semicircle perpendicular to the $x$-axis going through $p$ and $q$.
Best Answer
There are at least four "common" models of the hyperbolic plane:
The "upper" sheet of the hyperboloid of two sheets in Minkowski space (a.k.a., the set of future-pointing unit timelike vectors): $$ x_{1}^{2} + x_{2}^{2} - x_{3}^{2} = -1,\quad x_{3} > 0. $$ Hyperbolic lines turn out to be the intersections of planes through the origin with the hyperboloid.
The Klein disk model, viewed as the unit disk $$ x_{1}^{2} + x_{2}^{2} < 1,\qquad x_{3} = 1, $$ identified with the hyperboloid model by radial projection from the origin. Hyperbolic lines are Euclidean chords. (Patrick Ryan's Euclidean and Non-Euclidean Geometry is a good reference for these two models.)
The Poincaré disk model, viewed as the unit disk $$ x_{1}^{2} + x_{2}^{2} < 1,\qquad x_{3} = 0, $$ identified with the hyperboloid model by radial projection from the point $(0, 0, -1)$ (diagram below). Hyperbolic lines are arcs of Euclidean circles meeting the boundary of the disk orthogonally. Unlike the Klein model, the Poincaré model is conformal; hyperbolic angles coincide with Euclidean angles.
The upper half-plane model, also conformal, obtained from the Poincaré disk model via the fractional linear transformation $$ z \mapsto -i \frac{z + i}{z - i} = \frac{-iz + 1}{z - i}. $$ Hyperbolic lines are Euclidean semicircles (meeting the real axis orthogonally).
The Riemannian metrics in the Poincaré and upper half-plane models have well-known formulas in Euclidean coordinates $z = x + iy$: $$ ds^{2} = \frac{4(dx^{2} + dy^{2})}{\bigl(1 - (x^{2} + y^{2})\bigr)^{2}},\qquad ds^{2} = \frac{dx^{2} + dy^{2}}{y^{2}}. $$
Particularly, in the upper half-plane model, a Euclidean distance $ds = \sqrt{dx^{2} + dy^{2}}$ corresponds to a hyperbolic distance $ds/y$; as $y \to 0^{+}$, the hyperbolic length of a Euclidean segment of fixed length grows without bound.
Don Hatch has created an extensive gallery of hyperbolic tessellations (in the Poincaré model) that make this "length distortion" vivid. The "tiles" have fixed hyperbolic shape (and size), and their Euclidean representations shrink toward the boundary of the disk.
Another famous example is the Circle Limit series of prints by M. C. Escher. A web search for "Poincare disk" or "Poincare metric" should turn up many more diagrams.