The geometry in this probability question is unclear to me. On the circle: $x^2+(y-1)^2=1$ a point $B$ is chosen at random . Let $C(Z,0)$ be the point in which $0x$ axis and ray $AB, A(0,2)$ intersect. Find the distribution function of $Z$.
Answer:
Let abscissa of $D$ be $z$. Random variable $Z$ is less than $z$ if the point $B$ is on the arc $AE$ going counter clockwise. We have,
$$F_Z(x)=P\{Z<z\}=\frac{2 \pi - (\pi -2\arctan{\frac{z}{2}})}{2 \pi}= \frac{1}{2}+ \frac{\arctan{\frac{z}{2}}}{\pi}, z \in R$$
I don't know which angle to look at, and don't understand how they got this.
 A: The reesult depends on the definition of "chooosing of a point on the circle"!
I suppose that choosing a random point of the circle is equivalent to choosing a random angle $\alpha$ with uniform distribution over $[0,2\pi)$. As shown in the figure below, if $\alpha$ is given then
$$Z=2\tan\left(\frac{\alpha}{2}\right).$$

So, the distribution function of $Z$ can be calculated:
$$F_Z(x)=P(Z<x)=P\left(2\tan\left(\frac{\alpha}{2}\right)<x\right)=P\left(\alpha<2\arctan\left(\frac{x}{2}\right)\right)=\frac{1}{2\pi}\left[2\arctan\left(\frac{x}{2}\right)+\pi\right]$$
Here is the graph of the distribution function:

A: The distribution of $Z$ is also called "Cauchy Distribution".
Let the angle between AD and Y axis be $\theta$, $\displaystyle tan\theta = \frac{z}{2}$,
so $\displaystyle d\theta=\frac{2}{4+z^2}dz$.
$\theta$ can be any angle in the range of $\displaystyle (-\frac{\pi}{2}, \frac{\pi}{2})$, so $\displaystyle \theta \sim U(-\frac{\pi}{2}, \frac{\pi}{2})$,
and the density function is: $\displaystyle f(\theta)=\frac{1}{\pi}$.
Hence, here we get the cumulative distribution:
$\displaystyle F_Z(z)=\int^{\theta}_{-\frac{\pi}{2}}f(\theta)d\theta = \int^{z}_{-\infty} \frac{1}{\pi} \frac{2}{4+z^2}dz$ 
$\displaystyle =\frac{1}{2}+\frac{arctan\frac{z}{2}}{\pi}, \hspace{6mm} z\in R$ 
