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

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 $d=1$

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.