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What is the average distance between a (randomly chosen) point in a disc of radius r and a line segment of length $a < 2r$ whose midpoint is at the center of the disc? ["Distance" here being the shortest distance to any point on the line segment.]

We devote our calculations to only the first quadrant because all the other quadrants are symmetrical in terms of calculations.

Dividing the quadrant into three regions, we will use the concept of expectation value of a variable denoted as follows $$lt xgt = frac{int xP(x)dx}{int P(x)dx}$$
where $P(x)$ is the number of times a particular value $x$ of the quantity we seek occurs. (which we will contemplate as area of infinitesimal strips)

For region $mathbf I$:

The distances of all the points on a vertical line are the same ($x$).
So the $N^r$ contributed to the final formula of expected value or $N^r_{mathbf I}$ is calculated as $$int_0^{sqrt{R^2-frac{a^2}{4}}}xfrac a2dx=frac a4Biggl(R^2-frac{a^2}{4}Biggr)$$
where $frac a2dx$ is the area of the strip on which all the points are at a distance $x$ from the line of length $alt 2R$.
And obviously the denominator contributed $$D^r_{mathbf I} = area(rectangle OABC) =int_0^{sqrt{R^2-frac{a^2}{4}}}frac a2dx=frac a2sqrt{R^2-frac{a^2}{4}}$$
For region $mathbf {II}$:

$$N^r_{mathbf {II}}=int_{sqrt{R^2-frac{a^2}{4}}}^R xsqrt{R^2-x^2}dx$$
Subtituting $t=x^2$ and $dt=2xdx$, it becomes at last
$$-frac13biggl[(R^2-x^2)^{frac32}biggr]_{sqrt{R^2-frac{a^2}{4}}}^R=frac{a^3}{24}$$

where $sqrt{R^2-x^2}$ is the height of each strip.

And $D^r_{mathbf {II}}$ = area of half the segment $mathbf {II}$ $$= int_{sqrt{R^2-frac{a^2}{4}}}^R sqrt{R^2-x^2}dx$$
which can also simply be calculated by $frac12R^2(angle BOD)^c - area (triangle OBC)$
$$=frac12R^2sin^{-1}(frac a{2R})-frac a4sqrt{R^2-frac{a^2}{4}}$$.

For region $mathbf {III}$: ($mathbf {My attempt}$)

Since the equal distances between line and collection of random points are radial in sense, we now switch to polar coordinates.

Also you'd reconcile that there is some relation between $theta$ and the radial distance $r$ from point $A$ i.e. $max(r)$ depends on what $theta$ line you are seeing.
So, applying $mathit {Law of cosines}$ in $triangle OAE$:
$$cos(frac{pi}2+theta)=frac{r^2+frac{a^2}4-R^2}{ra}$$
or
$$sin(theta)=frac{R^2-big(r^2+frac{a^2}4big)}{ra}$$
or
$$r=frac{sqrt{4R^2-a^2cos^2{theta}}-asin{theta}}2$$
which is strictly decreasing in $big[0,frac{pi}2big)$.

Now, there are two integral sums that can lead us to $D^r_{mathbf {III}}$:
$$D^r_{mathbf {III}}=int_0^{R-frac a2}int_0^{frac{pi}2}rd{theta}dr+int_{R-frac a2}^{sqrt{R^2-frac{a^2}4}}int_0^{sin^{-1}biggl(frac{R^2-big(r^2+frac{a^2}4big)}{ra}biggr)}rd{theta}dr=frac{pi}4big(R-frac a2big)^2+int_{R-frac a2}^{sqrt{R^2-frac{a^2}4}}r {sin^{-1}Biggl(frac{R^2-big(r^2+frac{a^2}4big)}{ra}Biggr)}dr$$
or
$$D^r_{mathbf {III}}=int_0^{frac{pi}2}int_{R-frac a2}^{frac{sqrt{4R^2-a^2cos^2{theta}}-asin{theta}}2}rdrdtheta+int_0^{frac{pi}2}int_0^{R-frac a2}rdrd{theta}=int_0^{frac{pi}2}int_0^{frac{sqrt{4R^2-a^2cos^2{theta}}-asin{theta}}2}rdrdtheta$$
(I prefer latter to check if we are on the right track and we are! That's what I checked through above)

which can also simply be calculated by the area of half the segment $mathbf {III}$
$$=frac12R^2(angle BOA)^c - area (triangle OBA)=frac12R^2cos^{-1}Bigl(frac a{2R}Bigr)-frac a4biggl(sqrt{R^2-frac{a^2}4}biggr)$$
$$sum_{i=1,2,3} D^r_{mathbf {i}}= area(quadrant)=frac{pi}4R^2$$
since $sin^{-1}x+cos^{-1}x=frac{pi}2$

Since the distance between points at radial distance of $r$ from point $A$ is the same, thus multiplying $r$ in the integrand would mean $distancetimes (number of points having such distance)$ thing:
$$N^r_{mathbf {III}}=int_0^{frac{pi}2}int_0^{frac{sqrt{4R^2-a^2cos^2{theta}}-asin{theta}}2}r.rdrdtheta=frac1{24}int_0^{frac{pi}2}Big({sqrt{4R^2-a^2cos^2{theta}}-asin{theta}}Big)^3drdtheta$$

Average distance(as defined) of a random point from the given line is $$frac{underset{i=1,2,3}{sum} N^r_{mathbf {i}}}{frac{pi}4R^2}$$