Circle and a line!

There are 12 pairs ordered pairs of integers on the circle namely $(\pm1,\pm7), (\pm7,\pm1), (\pm5,\pm5)$ . We can draw a tangent on each of these points, so we have $n=12$ ordered pairs of $(a,b)$ for each of these tangents, since $ax+by=1$ represents all possible lines in the plane, except those of the form $y=mx$ and $x=0$ . Next, we can draw a line joining any $2$ of the $12$ points and since no $3$ points are collinear, all these lines will be distinct. So we have $\binom{n}{2}$ such lines and thus ordered pairs $(a,b)$ . However, any line joining a point $P$ and its reflection over the origin will pass through the origin and thus become of the form $y=mx$ , which isn't allowed. There are $n\over2$ such lines. So the final answer is $n+\binom{n}{2}-\frac{n}{2}=72$ .