Suppose I drew every chord in a circle. Is the chord/area ratio uniform or non-uniform?

I graphed a finite number of cords whose endpoints are nicely placed around a circle (a nice case).

This graph seems to suggest that there are more chords in certain regions than in others (for example, there are lots of chords near the end points and also in the center of the circle).

For lack of terminology, let's describe a chord 'density' as the number of chords around a very small rectangular region (in the image above for example, the density of any endpoint of a chord is 15).

Instead of a finite number of chords, let's say I drew every chord on a circle, then the density around any point would be infinity. But is the density uniform in the circle? That is, even if there are infinitely many chords, would some point have more chords than some other point? Sorry if I'm butchering the concept of infinity here.

If the density is not uniform, can we find a density function that describes the relative line density of any point in the circle?

Does a question like this even make sense? Does it make sense to describe a region's density if it's infinite?

This graph may illustrate what I mean by density (point to line segment, where density is highest at the bottom and increases to the left):

Edit: There may be a need to specify what a random chord is. In this case I define a random chord as a chord whose endpoints are randomly picked from the circle. Use infinitesimally small rectangular or circular (or other regions) as you see fit.

Sorry if the question is confusing. Please let me know what topic this should be under.

Solution 1:

I will choose a random chord by choosing two points on the circumference with uniform probability. You can then look at the sample space as the square $[0,2\pi) \times [0,2\pi)$. Each point in the square represents a chord and the selection is uniformly dense in the square. Now we can pick a point $A$ uniformly in the disk and ask what the chance a random chord intersects an $\epsilon$ disk around it, where $\epsilon$ is small so we can use small angle approximations. If the distance from the first point on the circumference to the point in the the disk is $d$, the angle subtended by the $\epsilon$ disk around $A$ is $\frac {2\epsilon}d$, so the arc of the circle that the second point can be in is $\frac {4\epsilon}d$ and the probability we hit the $\epsilon-$disk is $\frac {2\epsilon}{\pi d}$ This shows we need to average $\frac 1d$ over the set of first points. The density of chords around $A$ will be proportional to this.

We might as well use the unit circle centered at the origin and let $A$ be $(r,0)$. If the first point is at angle $\theta$ the distance to $A$ is $\sqrt{(r-\cos \theta)^2+\sin^2\theta}$. The average inverse distance to $A$ s then $\frac 1{2\pi}\int_0^{2\pi}\frac {d\theta}{\sqrt{(r-\cos \theta)^2+\sin^2\theta}}$ I couldn't get Alpha to do the integral for me but it seems chords will be densest around the edge.

I did a numerical integration using intervals of $4^\circ$ and find $$\begin {array} {r r} d&integral\\0&1\\0.2&1.01023144782371\\0.4& 1.04405634128953\\ 0.6&1.1145644874839\\0.8& 1.27024920049509\\0.9& 1.45186320070188\\ 0.95& 1.65255342453476 \end {array}$$ The values do not change much when I started at $2^\circ$ instead of $0^\circ$ so I think they are rather accurate. You can see the higher density near the edge. I suspect it will diverge as we get to $d=1$

Solution 2:

After Ross Millikan's answer $$f(r)=\frac 1{2\pi}\int_0^{2\pi}\frac {d\theta}{\sqrt{(r-\cos \theta)^2+\sin^2\theta}}=\frac 1{\pi}\left(\frac{K\left(-\frac{4 r}{(1-r)^2}\right)}{1-r}+\frac{K\left(\frac{4 r}{(1+r)^2}\right)}{1+r} \right)$$ where appear complete elliptic integrals of the first kind (this is valid for $r <1$).

Built around $r=0$, Taylor series give $$f(r)=1+\frac{r^2}{4}+\frac{9 r^4}{64}+\frac{25 r^6}{256}+\frac{1225 r^8}{16384}+\frac{3969 r^{10}}{65536}+\frac{53361 r^{12}}{1048576}+\frac{184041 r^{14}}{4194304}+O\left(r^{16}\right)$$ which seems to be quite good up to $r=0.7$ ($1.17478$ instead of the "exact" $1.17501$).

A much better approximation could be obtained using Padé approximants. For example $$f(r)=\frac {1-\frac{127285 }{96304}r^2+\frac{165891 }{385216}r^4-\frac{540191 }{24653824} r^6 } {1-\frac{151361 }{96304}r^2+\frac{20237 }{29632}r^4-\frac{1708091 }{24653824} r^6}$$ gives $1.44970$ for $r=0.9$ (the "exact" value being $1.45184$).

Below is a table (using the given Padé approximant) : $$\left( \begin{array}{ccc} r & \text{exact} & \text{approximation} \\ 0.00 & 1.00000 & 1.00000 \\ 0.05 & 1.00063 & 1.00063 \\ 0.10 & 1.00251 & 1.00251 \\ 0.15 & 1.00570 & 1.00570 \\ 0.20 & 1.01023 & 1.01023 \\ 0.25 & 1.01620 & 1.01620 \\ 0.30 & 1.02372 & 1.02372 \\ 0.35 & 1.03293 & 1.03293 \\ 0.40 & 1.04406 & 1.04406 \\ 0.45 & 1.05735 & 1.05735 \\ 0.50 & 1.07318 & 1.07318 \\ 0.55 & 1.09203 & 1.09203 \\ 0.60 & 1.11456 & 1.11456 \\ 0.65 & 1.14175 & 1.14175 \\ 0.70 & 1.17501 & 1.17500 \\ 0.75 & 1.21657 & 1.21655 \\ 0.80 & 1.27025 & 1.27016 \\ 0.85 & 1.34323 & 1.34283 \\ 0.90 & 1.45184 & 1.44970 \\ 0.95 & 1.64885 & 1.63250 \end{array} \right)$$

What is interesting is a look at the plot of $(1-r)f(r)$ for which a Pade approximant does a quite good job.