For $n$ dots, such that $n=10$:
pick a pivot, such that $p=[8, 11]$, calculate the distance to each:

let
let $r_{min} = $ smallest radius distance, which in all cases $0$
let $r_{max} = $ largest radius distance, which in this case $16$
Perform a circular expansion originated from $[8, 10]$ from $t_0 \rightarrow t_1$, where $t_0 = 0$ & $t_1 = 1$.
Both time and radius increment evenly, which
At $t_0$, $r_0 = r_{min} = 0$
At $t_1$, $r_1 = r_{max} = 16$
Question:
let
- quantity of dots be $n$
- vertexes of dots be $[[x_1, y_1], [x_2, y_2] ..]$
- pivot be $p = [x_p, y_p]$
- circular expansion starts at $t_0 = r_0 = r_{min}= 0$
- circular expansion end at $t_0 = 1$ & $r_1 = r_{max}$
Find $n(r)$, that $\int_{r_{min}}^{r_{max}} n(r) \,dr = n$
for the example above, here are some expected results:
- $\int_{0}^{2} n(r) = 2$
- $\int_{0}^{3} n(r) = 3$
- $\int_{0}^{7} n(r) = 7$
- $\int_{0}^{16} n(r) = 10$
The function should reflect in which segment of radius increment, there are relatively a large amount of dots getting covered, therefore, for any interval, for example, $n(6)$ the function should returns a decimal, perhaps approximately $6.7$
Thanks very much for reading to the end. I am not certain if I have spoken with enough clarity. It's a vague idea, I am not sure if it's tangible at all. I will be so excited if someone give me a hand.
We consider that the distances to the pivot point are ranked in increasing order, i.e., $r_k$ = distance to the $k$th closest point with $r_0=0$.
Your issue boils down to a linear interpolation between the following points :
$$(x,y)=(r_k, k)$$
(of course not in the same plane as the initial plane !).
Otherwise said, your function is affine on each interval $(r_k,r_{k+1})$ with:
$$\text{If} \ \ r_k \le x \le r_{k+1}, \ \ \ n(x)=\dfrac{(x-k)}{(r_{k+1}-r_k)}+k$$
Matlab program for the figure: