The Maximum number of points of intersection of 4 distinct circles and 8 distinct straight lines is
I will construct an explicit example. Take the following two configurations:
The second configuration is the eight-line solution to my generous lazy caterer problem. Now scale down the eight-line configuration so that all its intersections are inside the intersection of all the four circles (the small squarish region). If it happens that there is a multiple intersection, we can just tweak to remove it. Then we are guaranteed that every line intersects every circle twice, obtaining the maximum of $104$ intersections.
This is the result:
To generalise to any number of circles and lines:
- Set the sizes of the circles to be equal and make them encircle a point. This way every circle intersects every other circle twice, and there is a region inside all circles.
- Take a solution to the lazy caterer problem for the requisite number of lines – lines that all intersect each other – and scale the configuration down so that all intersections are inside the intersection of all circles. Generally, there is no multiple intersection, meaning that every line intersects every circle twice and we are finished. If there are multiple intersections, shifting the lines by a small amount (not enough to create multiple intersections elsewhere) suffices to remove the multiple intersection.