# Sum of smaller circles' area exactly equal to larger circle

I was making a thingy for my pond and needed to drill in a pipe a number of $1/2$" diameter holes more or less equal to the area of one $1.5$" diameter hole.

Through trial and error I discovered that the area of nine $1/2$ inch holes happens to be exactly equal to one $1.5$" hole. So then I wondered if this happens with other multiplies of $.5$ and discovered sixteen $1/2$" circles is exactly equal to one $2$" circle, twenty five to one $2.5$" circle, thirty-six to one $3$" circle, fourty-nine to one $3.5$" circle and sixty-four to one $4$" circle.

If there is a mathematical relationship, I don't see it, but I do find it interesting. If there is a formula for how many smaller circles equals the area of one larger circle, it would be useful. I'm guessing this would only happen if the smaller circle is an integer fraction of the larger circle.

The area of a circle is $\pi r^2$, so if you plan to fill up the area of a large circle (radius $R$, area $\pi R^2)$, you'll need $\left( \tfrac Rr \right)^2$ little circles. Just as you found empirically.
The area $A_R$ of a circle of radius $R$ is $\pi R^2$, so given two circles of radii $r, R$, the ratio of their areas is $$\frac{A_R}{A_r} = \color{red}{\frac{R^2}{r^2}}.$$ Notice that this can be an integer even when the ratio $\frac{R}{r}$ is not an integer, namely, when when $R = \sqrt{n} r$ for some positive integer $n$.