How can I calculate the minimum distance from the center to N points uniformly distributed in a circle? In other words, I want to find the distance to the closest point from the center. Does this distance decrease with N?
The distance from the center to random points inside a disk is of course random, so that your question doesn't quite make sense... You can compute its distribution, or its mean for instance.
As for the decrease, it is obvious that the minimum distance from the center to $\displaystyle N$ points decreases with $\displaystyle N$. Just give it a thought.
I show you how to find the distribution of the minimum distance from the center to $\displaystyle N$ points uniformly chosen in a disk.
First thing is to consider only one point $\displaystyle X$ uniformly distributed in a disk. I will consider a disk $\displaystyle D$ with center $\displaystyle O$ and radius $\displaystyle r$. Let $\displaystyle R=OX$ (i.e. $\displaystyle R$ is the distance from $\displaystyle O$ to $\displaystyle X$).
For $\displaystyle t>r$, we have of course $\displaystyle P(R<t)=1$, and if $\displaystyle 0<t<r$ then $\displaystyle P(R<t)=\frac{\pi t^2}{\pi r^2}=\frac{t^2}{r^2}$ (because $\displaystyle R<t$ means that $\displaystyle X$ is inside the circle with radius $\displaystyle t$, and the probability to be in a subset of the disk is proportional to its area). Using this computation, you can get the density of the distribution of $\displaystyle R$ (by differentiating, it is $\displaystyle \frac{2t}{r^2}$ on $\displaystyle [0,r]$ and 0 elsewhere), and the expectation of $\displaystyle R$.
Now, suppose there are $\displaystyle N$ independent points uniformly chosen on the same disk, and let $\displaystyle d$ be the minimum distance from the center to the points, i.e. $\displaystyle d=\min(R_1,\ldots,R_N)$ where $\displaystyle R_1,\ldots,R_N$ are defined like the previous $\displaystyle R$, for each of the points.
Then we have $\displaystyle P(d>t)=P(R_1>t,\ R_2>t,\ldots,\ R_N>t)=P(R>t)^N=\left(1-\frac{t^2}{r^2}\right)^N$, from which we can get the density of the distribution of $\displaystyle d$ : it is $\displaystyle \frac{2Nt}{r^2}\left(1-\frac{t^2}{r^2}\right)^{N-1}$ on $\displaystyle [0,r]$.
Sorry, my question was if N points are uniformly distributed in a sector of area $\displaystyle 1/2{r}^2\theta $, can i then write the pdf as $\displaystyle p(d>t) = \frac{Nt\theta}{\pi r^2}\left(1-\frac{\theta t^2}{2\pi r^2}\right)^{N-1}$ on [0,r]. Can I compute the expected value of d from the pdf by integrating over the area of sector ?(varying t from 0 to r and theta from 0 to thetamax - say pi/2). Also it doesn't matter (or does it ?) if d is minimum or maximum of R1, R2....RN ?. Please let me know.
Of course it does matter if $\displaystyle d$ is the min or the max! The min is small, it obviously tends to 0, while the max is big, it tends to $\displaystyle r$ when $\displaystyle N\to\infty$...
But you can find the distribution of the max using the same procedure as I did for the min: $\displaystyle P(d^*<t)=P(R_1<t,\ldots,R_n<t)=\cdots$, and deduce the expected value using for instance $\displaystyle E[d^*]=\int_0^r P(d^*>t) dt$.
Since the distribution of the points is rotation-invariant, this property is the same for any sector of radius $\displaystyle r$.