Find the maximum number of points of intersection of $m$ straight lines and $n$ circles.
This is a high school level counting problem and it has been answered on MSE before as I could find here.
And the answer turns out to be \begin{align} 2\binom n2+\binom m2+2n\cdot m \end{align}
But, isn't it that the number of possible points $\leq \left ( 2\binom n2+\binom m2+2n\cdot m\right)$? How do we know that such a configuration exists for all $n$?
Take circles with centres $(r,0)$ passing respectively through the points $(-2^{-r}, 0)$. It is easy to see that any two of these intersect in two points. The right-hand extremity of the larger of two of these circles lies outside the smaller, and the left-hand extremity lies inside it.
Now take any collection of lines in general position. Shrink it so that all the intersections fit inside the smallest circle and to the right of the line $x= 1$. The intersections of the lines are in the interior of all the circles, and the lines must therefore all cut each circle in two points.