How many circles (radius $r$) are needed to cover circle whose radius is $2$ times bigger (radius $2r$).
I think we need to use area which is $S=\pi R^2$ but I don't really know what to do.
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This is a disk covering problem. As you have stated it, it is not quite as difficult as some others.
The first task is to find the minimum number of small circles which cover the circumference of the bigger circle rather than the whole area. If this is $m$ then it will be impossible to have a regular $m$-gon with edges of length $2r$ fitting strictly inside the circle of radius $2r$. This implies $m \ge 6$.
It turns out to be just possible to cover the circumference of the bigger circle with 6 smaller circles, and ignoring symmetries there is only one way to do it (you get a regular hexagon whose vertices lie on the bigger circle and whose edges are diameters of the smaller circles). But this leaves an uncovered area in the middle, which needs at least one (and in fact exactly one) more.
So the answer is seven smaller circles.