Curvature of a hyperbolic plane Consider a projective plane and a real quadric.
According to the Klein-Beltrami-model the inside of the quadric is a  hyperbolic plane.
Klein proved that this plane has a constant negative curvature.   
Question 1: Is it possible to proof this in any easy way, i.e. without tensors or too much differential geometry?  
Question 2: Suppose I have an euclidean circle of radius R in a plane and I consider the inside of the euclidean circle as a hyperbolic plane. Is it possible to calculate the constant negative curvature of it (in relation to the value of R)?
EDIT for Question 2: for each value of $R$ every (negative) curvature $-a$ can be realised (see comment and answer A.D.Hwang).
I am looking for a mappping in the form of a formula:
Euclidean distance $\longleftrightarrow$ Hyperbolic distance
$R \longleftrightarrow \infty$
$0 \longleftrightarrow 0$
$x \longleftrightarrow y$
Is it perhaps the formula $y=\text{arccosh}{(1/ \sqrt{1 - (x/R)^2})}$?
If so, why does the curvature $-a$ not appear in the formula?
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$tl; dr:


*

*There's a hyperboloid model of the projective plane that is easily shown (using elementary linear algebra, see sketch below) to be homogeneous and isotropic under its isometry group. That's enough to guarantee its Gaussian curvature is constant. (In fact, every homogeneous surface has constant Gaussian curvature; isotropy isn't needed.)

*No, the curvature of a hyperbolic metric is independent of the Euclidean radius of a disk. For instance, if $a > 0$, the metric with components
$$
g_{11} = g_{22} = \frac{4a^{2}}{\bigl(1 - (x^{2} + y^{2})\bigr)^{2}},\quad
g_{12} = 0
$$
in the unit disk has Gaussian curvature $-a$.

In a nutshell, let
$$
H = \{(x, y, z) : -x^{2} - y^{2} + z^{2} = 1, z > 0\} \subset \Reals^{2, 1}
$$
be the set of unit future-pointing timelike vectors in Minkowski space, and let $g$ be the Riemannian (!) metric obtained by restricting the inner product
$$
\Brak{(u_{1}, v_{1}, w_{1}), (u_{2}, v_{2}, w_{2})}
  = u_{1} u_{2} + v_{1} v_{2} - w_{1} w_{2}.
\tag{1}
$$
For all real $\theta$ and $\alpha$, the set
\begin{align*}
e_{1} &= (\cosh\alpha \cos\theta, \cosh\alpha \sin\theta, \sinh\alpha), \\
e_{2} &= (-\sin\theta, \cos\theta, 0), \\
e_{3} &= (\sinh\alpha \cos\theta, \sinh\alpha \sin\theta, \cosh\alpha),
\end{align*}
is orthonormal with respect to the inner product (1). The first two vectors are spacelike, and the last is timelike (in fact, an element of $H$).
The general positive orthonormal basis is obtained by "rotating the $(e_{1}, e_{2})$-plane":
$$
(\cos\phi) e_{1} + (\sin\phi) e_{2},\quad
(-\sin\phi) e_{1} + (\cos\phi) e_{2},\quad
e_{3}.
\tag{2}
$$
The families of linear transformations with standard matrices
$$
\left[\begin{array}{@{}ccc@{}}
\cos\theta & -\sin\theta & 0 \\
\sin\theta & \phantom{-}\cos\theta & 0 \\
0 & 0 & 1
\end{array}\right],\qquad
\left[\begin{array}{@{}ccc@{}}
\cosh\alpha & 0 & \sinh\alpha \\
0 & 1 & 0 \\
\sinh\alpha & 0 & \cosh\alpha \\
\end{array}\right],
$$
namely rotations about the $z$-axis and boosts in the $(x, z)$-plane, act transitively on the set of positive orthonormal bases, and consequently generate the isometry group of $(H, g)$. (Each preserves both $H$ and the Minkowski inner product, hence induces an isometry. To show the action is transitive, start with an arbitrary positive orthonormal basis; rotate about the $z$-axis so that the timelike vector lies in the $(x, z)$-halfplane with $x > 0$; boost to make the third vector $(0, 0, 1)$; rotate about the $z$-axis to make the spacelike vectors $(1, 0, 0)$ and $(0, 1, 0)$.)
To show the curvature of $(H, g)$ is $-1$, it suffices to calculate at the point $(0, 0, 1)$. (Left to you.)
Hyperbolic lines are precisely the intersections of $H$ with a plane through the origin. (The intersection of $H$ with a plane containing the $z$-axis is the fixed-point set of a reflection isometry, hence a geodesic of $(H, g)$. Every plane through the origin that intersects $H$ can be brought to a plane containing the $z$-axis by a boost. Given points $p$ and $q$ in $H$, there is a unique plane containing $p$, $q$, and the origin; the intersection of this plane with $H$ is the geodesic from $p$ to $q$.)
The Klein model is the open unit disk in the plane $\{z = 1\}$, a.k.a., the image of $H$ under radial projection from $(0, 0, 0)$. A line is this model is a chord of the unit circle.
The Poincaré model is the open unit disk in the plane $\{z = 0\}$, a.k.a., the image of $H$ under radial projection from $(0, 0, -1)$.
