Average length of chord of circle - definite integral

Question

Given a circle C of radius a, and a diameter AB of C. Chords are drawn perpendicular to AB, intercepting equal arcs along the circumference of C. Find the limit of the average of the lengths of these chords, as the number of chords tends to infinity.

I can find the average lengths of the chords that intercept equal segments along AB: $$L_{ave}= 2*\frac{1}{a}\int_{0}^{a}\sqrt{a^2-x^2}\quad dx = 2*\frac{1}{a}* \frac{1}{4}\pi a^2 = \frac{\pi a}{2}$$

This is straightforward since $dx$ is invariable.

However, when solving the problem for chords intercepting equal arc segments, then $dx$ varies across the diameter of the circle. I was trying to set it up as a function of $sin$ or $cos$, but then I have both x and an angle as variables. Any hints on how to approach this?

Answer 1

Without loss of generality assume the circle is of radius 1 and divided by $n$ chords into $2n$ arcs. The required limit is then the Riemann sum $$\lim_{n\to\infty}\frac1n\sum_{k=0}^{n-1}2\sin\frac{(2k+1)\pi}{2n}$$ which is thus equivalent to the integral $$\int_0^12\sin\pi x\,dx=\frac4\pi$$ Therefore, for a circle of radius $a$, the average length is $\frac{4a}\pi$.

Answer 2

Depends which random variable you decide to average on. In one case you can average over the distance of the chord from the center which gives you R*pi/2 and in the other case you can average over the angle between the two extremities of the chord which gives you 4*R/pi.

One considers a uniform distribution of a distance between 0 to R (or - R to R) the other a uniform distribution of an angle between 0 to 2*pi.

Both distributions are not equivalent since there is a cos() transformation from one random variable to the other.