You can construct a solution like this:
Let Points be your n random points
foreach i in 1..(n-1):
foreach j in (i+1)..n:
construct the line L_{i,j} which has equal
distance to Points[i] and Points[j]
choose a random point c that is not on any of the constructed lines
this point is your center
You can guarantee for $c$ that for every circle around it, you will not have two points on the line. Why? Because if you had two points on the circle line, they would have the same distance in $c$ which means they would be on one line $L_{i,j}$. The same argumentation works for more than two points.
There are points left, because you've just drawn a finite number of lines which can't cover a plane.
As you can increase the size of the circle in steps, where you $r_t$ is the radius to the $t$-th next point, you will only have $t$ points in your circle.
Additional information
I don't know how a line of points that have equal distance to two Points $A, B$ is called in English (in German, it is "Mittelsenkrechte"). Here is an image of what I mean. The blue line is the "Mittelsenkrechte" for $A$ and $B$:
