What is the probability of the length distribution for lines cutting through a circle? I encountered this problem when I was reading a paper [1]. In it, the authors mentioned that if a circle has a radius of $c$, and a line cuts through it, then the probability for the length of this line lying between $\ell$ and $\ell+d\ell$ is
$\frac{\ell d\ell}{2c\sqrt{4c^2-\ell^2}}$ for $0<\ell<2c$ and $0$ for $2c<\ell$. I cannot understand how $\frac{\ell d\ell}{2c\sqrt{4c^2-\ell^2}}$ is derived. Could you please shed the light on me? Thanks!
[1] Cooper, M. G., B. B. Mikic, and M. M. Yovanovich. "Thermal contact conductance." International Journal of heat and mass transfer 12, no. 3 (1969): 279-300.
 A: There are many models for how a random line cutting through the circle can be chosen.
The one that the authors seem to be using is based on the idea that if I give you a point $P$ inside the circle, there is a unique line such that $P$ is its closest point to the center of the circle. (The line makes a chord inside the circle whose midpoint is $P$.) So the authors pick a uniformly random $r \in [0,c]$ and a uniformly random $\theta \in [0,2\pi]$, then draw the line whose closest point to the center of the circle is $(r \cos \theta, r \sin \theta)$.
To find the length of the chord, consider the right triangle through $O$ (the center of the circle), $P$ (the point above) and $X$ (one of the ends of the chord). Then $OX^2 = OP^2 + PX^2$ so $c^2 = r^2 + PX^2$ and $PX = \sqrt{c^2 - r^2}$. The length of the chord is twice $PX$: $2\sqrt{c^2-r^2}$.
To find the distribution of this quantity, we first find the CDF. The probability that the length of the chord is in $[0,\ell]$ is the probability that $2\sqrt{c^2-r^2} \le \ell$, or $r \ge \sqrt{c^2 - \ell^2/4}$. For $0<\ell<2c$, this describes an interval of length $c - \sqrt{c^2-\ell^2/4}$ for $r$, so the probability is $\frac{c - \sqrt{c^2-\ell^2/4}}{c} = 1 - \sqrt{1 - \frac{\ell^2}{4c^2}}$, because $r$ is uniformly chosen from $[0,c]$.
To find the probability that the length of the chord is in $[\ell,\ell+d\ell]$, we take the derivative of $1 - \sqrt{1 - \frac{\ell^2}{4c^2}}$, which gives us $-\frac1{2\sqrt{1 - \frac{\ell^2}{4c^2}}} \cdot -\frac{2\ell}{4c^2} = \frac{\ell}{2c\sqrt{4c^2-\ell^2}}$. After we multiply by $d\ell$, we get the answer in the paper.

Note, however, that different models for the random line will give different probabilities. My first guess about the random line was to pick two random points on the circle and draw a line through them. This gives the answer $\frac{2 d\ell}{\pi \sqrt{4c^2-\ell^2}}$, instead.
