A point is thrown at random inside a circle of radius r…

A point is thrown at random inside a circle of radius $r$; the probability of falling into a region inside the circle is proportional to its area. Determine the probability that:
$1^{\circ}$ the point is at distance at most $d$ $(d\lt r)$ from the centre;
$2^{\circ}$ the smaller angle between a given direction and the line joining the centre to the point does not exceed $\alpha$.

Step 1: $1^{\circ}$ — distance at most $d$

The points at distance $\le d$ from the centre form a disc of radius $d$ and area $\pi d^2$:

\[ \boxed{P_1 = \frac{\pi d^2}{\pi r^2} = \frac{d^2}{r^2}.} \]

Note: the answer is $(d/r)^2$, not $d/r$. For example, the chance of landing in the inner half of the radius ($d=r/2$) is only $\frac14$ — most of the area is near the edge.

Step 2: $2^{\circ}$ — angle at most $\alpha$

Let $\ell$ be the given direction (a ray from the centre). The "smaller angle" $\varphi$ between $\ell$ and the radius to the point takes values from $0$ to $\pi$. The condition $\varphi\le\alpha$ holds for the points of the sector bounded by the rays at angle $\alpha$ on both sides of $\ell$: a sector of angle $2\alpha$ and area $\frac{2\alpha}{2\pi}\pi r^2 = \alpha r^2$.

\[ \boxed{P_2 = \frac{\alpha r^2}{\pi r^2} = \frac{\alpha}{\pi}, \qquad 0\le\alpha\le\pi.} \]

The radius $r$ does not matter: all directions of the radius are equally likely, and $\varphi$ is uniformly distributed on $[0,\pi]$.

Step 3: Examples