What is the maximum number of points of intersection of 8 circles?

I came across this question on a book on combinations and permutations. Although it had a vaguely defined one step answer : $\textbf{$^8P_2 = 56$}$. As far as I could understand, the above step gives the number of arrangements possible for 8 objects taken two at a time. But I don't understand how it gives the points of intersection. Also, is there any other method to solve this kind of problems ? Regards

Two circles can intersect in at most two points. So if you have $8$ circles for every pair there can be at most $2$ intersections giving us the answer $2\cdot{8\choose 2} = 56$.

However this does in and of itself not show that this number can also be reached. There might after all are points where $ 3$ circles meet lowering the total number of intersections.

Fortunately it is easy to construct $8$ circles that have no triple intersections. For example you can place $8$ circles all with radius $8$ on the coordinates $(0,0), (1,0),\ldots,(7,0)$. All these circles clearly intersect twice with each other. But if there was some point $x$ where three circles intersected then the circle with radius $8$ around $x$ would intersect the $x$-axis in $3$ points, which is clearly absurd.

Two circles can intersect in at most two points (excluding the case where two circles coincide with each other). So by looking at all possible pairs of circles, the maximum number of intersection points is given by $2\binom{8}{2}=56$. (A simple proof of the need for the multiplier of two is that two circles can intersect in 2 points, whereas $\binom{2}{2}=1$).

Now consider the case where all circles are the same size and close enough together that the center of each circle lies inside every other circle. In the general case where none of the pairwise intersection points lies on the perimeter of any of the six other circles, the maximum number of intersection points is realisable.