Counting intersection points of circles and lines

Question

Find maximum number of points of intersection of 7 straight lines and 5 circles when 3 lines are parallel and 2 circles are concentric.

My attempt: Total intersection points = Total intersection points of lines with lines ($T_{ll}$) + Total intersection points of circles with circles ($T_{cc})$ + Total intersection points of lines with circles $(T_{lc})$.

$T_{ll}$ = ${7}\choose{2} $ - ${3}\choose{2}$ = 18. Each pair of lines intersect once except for the three parallel lines.

$T_{cc}$ = $2\cdot {4\choose2}$ + $2\cdot {4\choose 2}$ = 24. The three non concentric circles intersect with each of the two concentric circles at 2 points each.

$T_{lc}$ = $ 5 \cdot 7 \cdot 2$ = 70 . Each line can intersect w/ each circle twice at max.

Total = 112

But the answer in the book is given 106.

I think i might have gone wrong in calculating $T_{lc}$ or $T_{cc}$ but I am not sure.

I am not very good at combinatorics (as you can probably see) so I would really appreciate it if someone could tell me not just the solution but the thought process behind the solution too and also where i went wrong. Also if you have some tips on how to tackle these kinds of problems and combinatorics in general, do let me know.

Thanks!

Answer 1

Your $T_{cc}$ is overcounting because in first $2\cdot {4 \choose 2}$ you already counted the intersection points of non-concentric circles. In second $2\cdot {4 \choose 2}$, you're still counting these. Correction would be $$2 \cdot {4 \choose 2} + 2\cdot 3=18$$

You can also find $T_{cc}$ in following, slightly more illustrating manner. Two circles can intersect in maximum two points. Draw two concentric circles. Now introduce third circle. This gives $4$ intersection points, two with each of previous ones. Fourth circle gives $6$ more points. Fifth gives $8$ more points. So $$T_{cc}=4+6+8=18$$

Answer 2

$7$ lines can intersect in $\dfrac{7\cdot6}2=21$ points. But if $3$ are parallel, you deduce $\dfrac{3.2}2=3$.

$7$ lines and $5$ circles can intersect in $2\cdot3\cdot5=70$ points.

$5$ circles can intersect in $2\dfrac{5\cdot4}2=20$ points but if $2$ are concentric, you deduce $2$.

Anyway, this is an upper bound, there is no guarantee that all are possible simultaneously.