In a plane, a point A and the centre B of a circle of radius…

In a plane, a point $A$ and the centre $B$ of a circle of radius $R$ move independently along straight lines with constant speeds $u$ and $v$. At a given moment $AB=r$ $(r\gt R)$, and the angle between the line $AB$ and the velocity $\vec v$ of the circle is $\beta$. Assuming that all directions of motion of $A$ are equally likely, find the probability that the point $A$ enters the circle.

Step 1: Relative motion

In the frame of the circle, the circle is fixed and $A$ moves with velocity $\vec w=\vec u-\vec v$. $A$ enters the circle iff the ray from $A$ in the direction of $\vec w$ meets the disc. Seen from $A$, the disc occupies the angle

\[ |\,\text{angle between } \vec w \text{ and } \overrightarrow{AB}\,|\lt\gamma, \qquad \gamma=\arcsin\frac Rr. \]

Step 2: The stationary circle ($v=0$)

Then $\vec w=\vec u$ has a uniform direction, and the favourable directions form an angle $2\gamma$ out of $2\pi$:

\[ P = \frac{2\gamma}{2\pi} = \frac1\pi\arcsin\frac Rr. \]

Step 3: The moving circle, $u\gt v$

Put $\overrightarrow{AB}$ along the $x$-axis. Let $\varphi$ be the direction of $\vec w$ and $\theta$ the direction of $\vec u$. Resolving $\vec u=\vec w+\vec v$ along and across the direction $\varphi$: the component of $\vec v$ across $\varphi$ is $v\sin(\beta-\varphi)$, so

\[ \theta = \varphi+\arcsin\left(\frac vu\sin(\beta-\varphi)\right). \]

For $u\gt v$ this is an increasing one-to-one correspondence between $\theta$ and $\varphi$ (every direction $\varphi$ of $\vec w$ is reached exactly once). The favourable $\varphi$ are $(-\gamma,\gamma)$, so the favourable $\theta$ form an interval of length $\theta(\gamma)-\theta(-\gamma)$, and since $\theta$ is uniform on a circle of length $2\pi$:

\[ \boxed{P = \frac{1}{2\pi}\left[2\gamma+\arcsin\left(\frac vu\sin(\beta-\gamma)\right)-\arcsin\left(\frac vu\sin(\beta+\gamma)\right)\right],\qquad \gamma=\arcsin\frac Rr.} \]

Step 4: Checks

• $v=0$: $P=\frac\gamma\pi$, as in Step 2 ✓.
• $\beta=\pi$ (the circle moves towards $A$): $P=\frac1\pi\big[\gamma+\arcsin(\frac vu\sin\gamma)\big]\gt\frac\gamma\pi$ — easier to hit.
• $\beta=0$ (the circle moves away from $A$): $P=\frac1\pi\big[\gamma-\arcsin(\frac vu\sin\gamma)\big]\lt\frac\gamma\pi$ — harder.
• Simulation with $u=2$, $v=1$, $r=5$, $R=1$, $\beta=0.7$: $0.0382$ against the formula $0.0384$.

Step 5: The case $u\lt v$

Then the circle of possible relative velocities (radius $u$, centre $-\vec v$) does not surround $A$: only directions within $\arcsin\frac uv$ of $-\vec v$ are possible, and some of them are reached twice. The probability is still "the fraction of the circle of relative velocities that lies inside the cone of hits", but it must be computed case by case; in particular, if the cone of hits and the cone of possible relative velocities do not overlap, then $P=0$.