What is this geometric Probability 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 $c$? 

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!
