Bertrand's Paradox and uniform distribution This is a problem from Hans-Otto Georgii's textbook.
Recall the situation of Bertrand’s paradox, and let X be the distance of the random chord to the centre of the circle. Find the distribution density of X if
(a) the midpoint of the chord is uniformly distributed on the disk $|x|<r$,
(b) the angle under which the chord is seen from the centre of the circle is uniformly distributed on the interval $\Omega_2$ = $[0,\pi]$.
The solutions are given as:
a. the distribution density of X on [0,r] is $\rho_1(x)$ = $2x/r^2$.
b. $\rho_2(x) = 2/(\pi*\sqrt(r^2 - x^2))$, x $\in [0,r]$.

edit:
the overall approach is to derive cdf then differentiate.
a. cdf = $\pi x^2/ \pi r^2$; pdf = $2x/r^2$ .
my thoughts for b: 
The probability formula for continuous uniform distribution in my textbook:
$U_\Omega(A) = \lambda^n(A) / \lambda^n(\Omega)$. So I tried expressing cdf like some probability:
$\Omega = [0,\pi]$, so $\lambda^n(\Omega) = \pi$. A is the range of angle POQ when the distance is in $[0,x]$. PQ is the chord.
A = $2arccos(x/r)$ for x in $[0,x]$. 
cdf = $2arccos(x/r) / \pi$; pdf = $- 2 / \pi \sqrt(r^2 - x^2)$.
However there is an extra negative sign here. could anybody help correct my workings?
 A: Although the question is somewhat older, let's derive the solution to both parts in more detail than provided in the book (and comments).
In the end, the exercise consists of basic trigonometry and calculus together with some probability fundamentals.

Solution to (a).
Consider the following image created in Tikz which shows some circle $Z$ and a (uniformly drawn) random point within the circle.

The distance of the circle center to that point is $x$. The set of points $A$ that have a distance $\leq x$ from the circle center is $A := \{y \in Z: \Vert y \Vert \leq x\}$ and is hatched blue. In other words, all points of $A$ lie in a smaller circle with radius $x$ around the circle center of circle $Z$. Since $\vert A \vert = \pi x^2$ and $\vert Z\vert = \pi r^2$, we have that
\begin{equation}
F_X(x) = P(X \leq x) = \frac{\vert A \vert}{\vert Z\vert} = \frac{\pi x^2}{\pi r^2} = \Big(\frac{x}{r}\Big)^2
\end{equation}
as you already edited into your question.
But then
\begin{equation}
\frac{\text{d}}{\text{d}x} F_X = \frac{2x}{r^2} 
\end{equation}

Solution to (b).
Consider the following image, also created in Tikz.

Let $r$ be the radius and $X$ the distance of the circle center to the midpoint $M_1$ of the chord.
Based on the image and basic trigonometry, it is clear that $\frac{\alpha_1}{2} = \arccos(\frac{X}{r})$. Conversely, it is $X = \cos\Big(\frac{\alpha_1}{2}\Big)r$.
The probability for a distance $X \leq x$ with $x \in [0, r]$ is hence
\begin{equation}
 F_X(x) = P(X \leq x) = P(\cos\Big(\frac{\alpha_1}{2}\Big)r \leq x) = P(\alpha_1 \geq 2\arccos(\frac{x}{r})) = 1 - P(\alpha_1 < 2\arccos(\frac{x}{r})) = 1 - \frac{2}{\pi}\arccos\Big(\frac{x}{r}\Big)
\end{equation}
and the probability density therefore
\begin{equation}
 f_X(x) = \frac{\text{d}}{\text{d}x} F_X(x) = \frac{2}{\pi}\frac{1}{\sqrt{1 - \Big(\frac{x}{r}\Big)^2}}\frac{1}{r} = \frac{2}{\pi}\frac{1}{\sqrt{r^2 - x^2}}
\end{equation}
since $r > 0$.
