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In a circle of radius $R$ two points are chosen at random(the points can be anywhere, either within the circle or on the boundary). For a fixed number $c$, lying between $0$ and $R$, what is the probability that the distance between the two points will not exceed $c$?

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Wlog, you can fix one of the two points, then how is the distribution of the lenque of the chord? –  Lost1 Jan 5 at 15:45
The point (you are fixing)does not necassarily lie on the boudary;it can be inside the circle as well –  sajjad veeri Jan 5 at 16:00
Please define how you want to choose a point in the circle randomly. –  Michael Hoppe Jan 5 at 16:16

2 Answers 2

This is not a complete answer to the problem, but a rephrasing of the situation which may help. Take the disk and draw a diameter (chord passing through the center). Now choose, as you say, two points at random in the disk and draw a segment connecting the points. Rotate the disk so that the diameter is parallel to this segment. Now we can rephrase the situation as follows:

Pick two points at random from an interval of length $2R$. What is the probability that the length of the interval with these endpoints does not exceed $c$? I think this question is a bit easier to answer.

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Unfortunately I don't have enough time to write down the whole thing. Basically you have to distinguish two cases: A circle with radius c around the first point lies completely inside the circle that is span by R. Then the probability is $$\frac{c^2}{R^2}$$, but this only happens in $$\frac{(R-c)^2}{R^2}$$ of the cases.

Therefore the total probability will be $$\frac{c^2\cdot(R-c)^2}{R^2}+K\cdot\left(1-\frac{(R-c)^2}{R^2}\right)$$

With $K$ being the probability of the distance between the two points larger then c in case of the circle around the first point with radius c intersecting with the outter circle. And $\left(1-\frac{(R-c)^2}{R^2}\right)$ being the probability for this to happen.

In order to calculate K you need the formula for circle-circle intersections:

Given two circles with radi R and r the formula for the intersectionarea A is:$$A(R,r,d)=r^2\cos^{-1}\left(\frac{d^2+r^2-R^2}{2dr}+R^2\right)\cos^{-1}\left(\frac{d^2+R^2-r^2}{2dR}\right)-\frac{1}{2}\sqrt{(-d+r+R)(d+r-R)(d-r+R)(d+r+R)}$$

Please make sure I don't have any typos, the original formula can be found here: wolfram

The $d$ in the formula is the distance between the two circle's centers.

We will now integrate over this $d$.

$$K = \int_{d=R-c}^{d=R}\left(A(R,r,d)\cdot\frac{d^2}{R^2}\right)$$

You have to multiply with $\frac{d^2}{R^2}$, since the probability grows with $d$ that the first point picked lies at a specific distance from the center. If you plug this into mathematica or something similar you should have your result!

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