Assume $n$ points $P\_i \in \mathbb{R}^2, i \in {1,2,...,n}$. For each point there is a $k$ nearest neighbour $(k<n)$, or equivalently for each point $P\_i$ there is one circle with center the point $Pi$ and radius $r\_i$ such that the circle contains exactly $k+1$ points (considering also the center $Pi$). The question is what is maximum number of circles that a point can participate? 


For one nearest neighbor ($k=1$) and by using the properties of regular hexagon I can prove that the maximum number is 5 (less than 6). Is there any way to extend it for $k>1$?



k nearest points
What you're asking is: what is the ply of the system of $k$-nearest-neighbor balls? Here the ply is the maximum number of balls that have a common intersection. It's not quite the same as the degree of the $k$-nearest-neighbor graph (Fischler's answer) because the ply can be maximized at a point that's not one of the given ones.


Anyway, it is known that in any dimension the ply is at most $\tau k$, where $\tau$ is the kissing number; this is only slightly weaker than the bound given by Fischler's answer. See e.g. Separators for sphere-packings and nearest neighbor graphs, Miller et al., JACM 1997 (p.16 of [the preprint version](https://www.cs.bu.edu/faculty/steng/teaching/Fall2005/cs532/jacm.pdf)). 


