The Maximum number of points of intersection of 4 distinct circles and 8 distinct straight lines is The Maximum number of points of intersection of 4 distinct circles and 8 distinct straight lines is
1)66
2)64 
3)104
4)40
Can anyone please help me to solve this problem?
My attempt : I have seen this link--Maximum number of points of intersection. But I can not understand how they are so sure that all intersection points are distinct. 
 A: 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.

A: The trick is that you can place the individual objects (circles + lines) an arbitrarily tiny (but nonzero) distance/angle apart. Then maximum intersection is guaranteed as long as you make sure that there are no multiple intersections. For example, proceed as follows:


*

*Draw an orientation point and make sure that it is inside of all the circles. Think of a thin pole where you drop your 4 rings over. Then every ring (circle) must necessarily have at least two intersection points with every other ring. 

*Now you have to avoid multiple intersections. Since every added circle just has to avoid a finite set of previous intersection points, that is easy to achieve: just avoid them when fine-tuning the ring positions. You have continuous area between the other points. ;-)

*Now there is a small area around the center pole that is part of all circles (the intersection area of all the circles)

*Take infinitely long spaghettis (lines) and drop them randomly. As long as neither two of the spaghettis are in parallel, they will intersect maximally (one point for each pair). If they are in parallel, adjust them. If multiple intersections occur, adjust as well. As in (2) you have plenty of space for the adjustment due to continuity between the previous intersection points.

*Finally you have to make sure that all spaghetties go through the interior of all the circles. This is the case, if at least one point of every line is inside the intersection area of all the circles. It is enough to make sure that all the spaghetti-to-spaghetti intersections are inside the intersection area (3) of all the circles. Since the former is a finite set, you can scale and move your spaghetti configuration accordingly until they are all inside.


Et voila: you have reached the number of intersections that are combinatorially possible under the given requirements.
A: An interesting way to understand the 8-line situation is with inversion. As we know a circle that passes through the center of inversion is inverted into a line. Choose the original as the center of inversion, and choose 8 evenly spaced points on the unit circle with angles ranging from 0 to $\frac{7\pi}{8}$ as centers and draw 8 circles passing through the origin, we easily get 8 circles that intersect with each other at distinctive points. Invert the 8 circles centered at the origin, we get 8 lines intersecting each other at distinct points.
8 circles inverted into 8 lines
