So many circles...

Each circle can intersect another circle a maximum of two times. So avoiding double counting, the maximum number of points of intersection with $n$ circles is $2\times T_{n-1} = n\times (n-1)$ where $T_n$ is the $n$ th triangular number. So this is just a problem of counting trailing zeros now. $1000000000$ has $9$ trailing zeros, so the answer will have $9$ nines followed be $9$ trailing zeros.

all ya got to do is n(n-1) where n is the number of circles