Intersectception

There would be $n \times (n-1)$ intersections to $n$ circles. These will create $n \times (n-1)$ distinct areas, excluding the portion common to all the $n$ circles and also excluding the exterior.

Therefore, $a_{10} = 10 \times 9 + 2$ $92$

This problem can be solved also by using induction and seeing straightforward in what way should the intersection of that circle behave such that it is assured that it generates the maximum number of regions. Observe that for a number of circles which will have a maximum number of regions already generated adding one other circle will necessary intersect with 2 circles in a region which is just common for those circles and then intersect the next circle in a region which must belong also to that circle and one of the first circles intersected by the new circle ,and that afterwards it must intersect the 3nd circle in a region which is common with the other 2 circle in which it has intersected and so for 4 on until the region which has all the regions common which will generate also 1 more region. Taking into account also the generated region of that circle added this will give that the number of regions added to the already number of regions will be for some number n 2(n-1). Therefore for each step of the induction it can be interpreted that the number of regions added as the number of circles grows will grow by a number of 2 and therefore that the number of circles generated imply if seen as such a relation a pretty simple kind of exponential growth of the kind 1+1+2 *1+2 *2+2 *3+2 *4 and so on which applied gives 92 or so I think cause it is a bit confusing anyways.