5. Nine pairwise noncongruent circles are drawn in the plane such that any two circles intersect twice. For each pair of circles, we draw the line through these two points, for a total of $\binom{9}{2}=36$ lines. Assume that all 36 lines drawn are distinct. What is the maximum possible number of points which lie on at least two of the drawn lines?
Answer:
462
The lines in question are the radical axes of the 9 circles. Three circles with noncollinear centers have a radical center where their three pairwise radical axes concur, but all other intersections between two of the $\binom{9}{2}$ lines can be made to be distinct. So the answer is
$$
\left(\begin{array}{c}
9 \\
2 \\
2
\end{array}\right)-2\binom{9}{3}=462
$$
by just counting pairs of lines, and then subtracting off double counts due to radical centers (each counted three times).