Greatest number of regions we can get when dividing with lines and circles

Lines

For a plane with $n$ lines, consider what happens when you add a line. It divides all the regions through which it passes into two, thus adding one region for each region it passes through. The number of regions it passes through is the number of lines it crosses plus one, that is the number of points it creates plus one for itself. Thus, the number of regions added to the original one, $\binom{n}{0}$, is the number of points, $\binom{n}{2}$, plus the number of lines, $\binom{n}{1}$. Thus, the maximum number of regions is $$ \binom{n}{0}+\binom{n}{1}+\binom{n}{2}=\frac{n^2+n+2}{2} $$

Circles

For a plane with $n$ circles, consider what happens when you add a circle. It divides all the regions through which it passes into two, thus adding one region for each region it passes through. The first circle divides the plane into two. The number of regions it passes through is at least one, if it doesn't intersect any other circles, and up to the number of crossings with other circles. For $n$ circles there can be up to $2\binom{n}{2}$ crossings. Thus, the first circle divides the plane into two, $2\binom{n}{0}$, then the rest of the circles add the number of crossings. Thus, the maximum number of regions is $$ 2\binom{n}{0}+2\binom{n}{2}=n^2-n+2 $$

if plane divided by lines, recursion function is $R(n+1)=R(n)+n+1$, where $R(0)=1$, $n\geq0$ Solving recurrence $R(n)=\frac{n^2+n+2}{2}$
if plane divided by circles, recursion function is $R(n+1)=R(n)+2n$, where $R(1)=2$, $n>0$ Solving recurrence $R(n)=n^2-n+2$