Curvature of curves on the hyperbolic plane In Euclidean geometry the curvature of a line or curve in the reciprocal of the radius of the kissing circle 
In hyperbolic geometry we cannot use this as measure of curvature (what is the curvature of an hypercycle?) 
What measure can we use instead as measure of curvature?
 A: The natural notion to use is Geodesic curvature which makes sense for curves on any Riemannian manifolds. The name comes from the fact that geodesics have zero curvature. 
For example, on the hyperbolic plane with Gaussian curvature $-1$, horocycles have geodesic curvature $1$. Indeed, let's consider Poincaré half-plane model with metric $(dx^2+dy^2)/y^2$. The line $y=1$ is a horocycle that is naturally parametrized by arclength, $\alpha(t) = (t, 1)$. The tangent vector is the unit vector that points to the right. This makes it appear as if the horocycle doesn't curve but  we should use parallel transport to judge whether two vectors are parallel.  
Take two points $A=(t, 1)$ and $B=(t+h, 1)$ on the horocycle and draw a geodesic between them: it's an arc of a circle with Euclidean radius $\sqrt{1+(h/2)^2}$. The tangent vector to $\alpha$ makes the angle $\sin^{-1}(h/2)$ with the geodesic at both $A$ and $B$, but it's in opposite directions. So, transporting the vector $\alpha'(t)$ from $A$ to $B$ along the geodesic, we see that at the point $B$ it makes the angle $2\sin^{-1}(h/2)$ with $\alpha'(t+h)$. Since the unit tangent rotates by $2\sin^{-1}(h/2)$ over the distance $h$, the geodesic curvature is 
$$
\lim_{h\to 0} \frac{2\sin^{-1}(h/2)}{h} = 1
$$
Disk model
On second thought, it's easier to use the disk model; the metric will be $4ds^2/(1-x^2-y^2)^2$ so the curvature is still $-1$. The diameter $(-1,1)\times \{0\}$ is a geodesic, and near the center $(0,0)$ its arclength parameterization moves approximately as $t\mapsto (t/2,0)$ when $t\approx 0$. So the parallel transport along this geodesic for small distances near center will be Euclidean, which implies that the geodesic curvature of any curve tangent to this geodesic at $(0,0)$ will be just $1/2$ of its Euclidean curvature at that point. (Here $1/2$ comes from the aforementioned speed of parameterization). 
Summary: to compute geodesic curvature in the hyperbolic disk model, move the point of interest to the center by a Möbius transformation, and take $1/2$ of Euclidean curvature there. Examples:


*

*Horocycles have geodesic curvature $1$, as shown by the horocycle $x^2+(y-1/2)^2=1/4$ that passes through $(0,0)$ and has Euclidean curvature $2$.

*A hyperbolic circle of hyperbolic radius $R$ has geodesic curvature $1/\tanh R \in (1,\infty)$. Indeed, when such a circle passes through $(0,0)$, its point furthest from $(0,0)$ is at hyperbolic distance $2R$ from $(0,0)$. Solving $2\tanh^{-1} d = 2R$ yields $d=\tanh R$ for the Euclidean diameter of this circle, so its Euclidean curvature is $2/\tanh R$ and geodesic curvature is $1/\tanh R$

A: Using Hyperbolic geometry to calculate the geodesic curvature of a circle
If the circle has centre M and a radius $r$ 
Imagine 2 points on the circle $A$ and $B$
let $C$ be the midpoint of the segment (chord) $AB$
Then the triangle $\triangle ACM$ is a right angled triangle (right angle is $\angle ACM$) and $\angle M = \angle AMC = 1/2 \angle AMB $
For this triangle we have the following formula's


*

*$ \tan(\angle CAM)   = \cot(\angle M) / \cosh(r) $

*$ \sinh(AC) = \sin(\angle M)   \sinh(r) $

*$ \tanh(MC) = \cos(\angle M)   \tanh(r) $ (not needed)


The angle between the tangent at A and $AB$ is
$$ \frac{\pi}{2} -\angle CAM = \frac{\pi}{2} -\arctan \left( \frac{\cot(\angle M)}{\cosh(r)} \right) = \arctan \left( \tan ( \angle M)\cosh(r) \right) $$
And the curvature is $ \lim_{\angle M \to 0} \frac{\arctan \left( \tan(\angle M)\cosh(r)\right)} {\sin(\angle M)   \sinh(r)} $
Because $ \lim_{\angle M \to 0} \frac{ \arctan \left( (\tan(\angle M) C \right) }{\sin(\angle M)} = C $
The limit is  $\frac{\cosh(r)}{\sinh(r)} =\frac{1}{\tanh(r)}$
A: A useful notion of curvature in other spaces is the Geodesic Curvature $k_{g}$. We will find the geodesic curvature of a hyperbolic circle.
One way of modelling a hyperbolic cicle is by defining it as the boundary of a ball in the Poincare Unit Disk Model of the Hyperbolic Plane.
First one defines the Poincare unit disk as:
$\mathbb{D}^2 = \{z \in \mathbb{C} ; |z| \leq 1 \}$, equipped with co-efficients of the first fundamental form $(E=G=\frac{1}{(1+|z|^2)^2},F=0)$. One can use these co-efficients to calculate the Gauss Curvature of the hyperbolic plane to be $K = -1$.
And the Ball of radius R as:
$B = \{ z \in \mathbb{C} ; |z| \leq \tanh(R/2) \}$
The boundary of this ball $\partial B =  \{ z \in \mathbb{C} ; |z| = \tanh(R/2) \} $ is the circle we are concerned with. We can see this is a circle since if we define the distance $d$ between any two points in this space as the geodesic then $d(0,e^{i\phi}\tanh(R/2)) = R$, which is a constant.
If $w = w_1+iw_2 \in T_{z}\mathbb{D}^2$ then the norm of $w$ is:
$||w||_{z} =(w_1 , w_1) \begin{pmatrix}
E & F \\
F & G  \\
\end{pmatrix} \begin{pmatrix}
w_1 \\
w_2  \\
\end{pmatrix}$
If one were to parameterise $\partial B$ with the curve $\alpha(t) = \tanh(R/2)e^{it}$, then you would find $||\alpha''(t) = constant||$, which implies that the geodesic curvature is constant.
Now we put everything together using the Gauss-Bonnet Theorem, Which states (in our case with no interior angles and an Euler Charecteristic of $1$):
$\int_{\partial B}k_{g} ds + \int_{B}KdA = 2\pi $
This is simply (by use of constant geodesic curvature and gauss curvature of $-1$):
$k_{g} \int_{\partial B}ds - \int_{B}dA = 2\pi $
The integral over the boundary is just the perimeter of the circle in this space, which can be found by our parameterisation and the calculating the norm of the derivative of our parameterisation. This yields $Perimeter = 2\pi \sinh(R)$
The integral over the whole ball is just the area of the ball, which can be calculated by the co-efficients of the first fundamental form, yielding $Area = 4\pi \sinh^2(R/2)$.
Plugging in the formulae for perimeter and area,re-arranging for $k_{g}$ and using various trigonometric identities we get:
$k_{g} = \frac{1}{\tanh(R)}$
