Answer Choices- 16, 12, 8, 6

I believe it is 12 .   That is what I find.

Each pair of circles intersect twice

So.....we have a set of n circles and we want to choose any two of them  = C (n, 2)

And since each pair intersects twice, the number of intersection points = 2 C(n , 2)  =

2 n! / [ (n - 2)! * 2! ]   =  n! / ( n - 2)!  =   n ( n - 1)

So.....The max intersection points of n circles =  n(n - 1)

So.....  4(4 - 1)  =  4 (3)   =  12

Just as EP found !!!!