Thinking it is fun

Case-1: Two straight lines intersect

Since two straight lines can intersect at only one point, maximum possible number of such points $=\binom{8}{2}=56$

Case-2: Two circles intersect

Since two circles can intersect at maximum of 2 points, maximum possible number of such points $=2×\binom{4}{2}=12$

Case-3: One circle and one line intersect

Since one circle and one line can intersect at maximum of two points, maximum possible number of such points $=2×\binom{8}{1}\binom{4}{1}=64$

Adding up, we get $64+12+28=\boxed{104}$

Case-1: Two straight lines intersect Since two straight lines can intersect at only one point, maximum possiblenumber of such points Case-2: Two circles intersect Since two circles can intersect at maximum of 2 points, maximum possiblenumber of such points Case-3: One circle and one line intersect Since one circle and one line can intersect at maximum of two points, maximum possible number of such points Adding up, we get 64+12+28 = 104

Max no of points by the intercestion of n circles and m straight lines is nP2+mC2+2nm nP2 gives the no of intersections of circles mP2 gibes no of intersection of straight lines 2mn gives the intersection of lines with circle since each straight line cuts a circle at two points Hence , 4P2+8P2+2 8 4 gives us the max intersections I.e., 12+28+64=104