For the two attached images herewith, let $P_n$ denote the proportion of the big circle covered by the small circles as function of the number of the small circles in the outer layer. Find the expression for $P_n$ and compute the $$\lim_{n \to \infty} P_n$$
Author's answer to this question as follows: Let R be the radius of the big circle, $r_n$ be the radius of the small circles in the outer layer, and $R_n$ be the radius of the circle encompassing all small circles except those in the outer layer. Then in both cases $P_n R^2\pi = n r^2_n \pi + P_n R^2_n \pi (\star)$ and $\sin{\frac{\pi}{n} }= \frac{r_n}{R - r_n}$ In the first case, $R_n= R- 2r_n$, giving $$ P_n = \frac{nr_n}{4(R-r_n)} = \frac{n \sin{\frac{\pi}{n}}}{4}, \lim_{n \to \infty} =\frac{\pi}{4}$$
In the second case we have $(R-r_n) \cos{\frac{\pi}{n}}- \sqrt{(r_n + \hat{r}_n )^2 - r^2_n} + \hat{r}_n = R_n $ where $\hat{r}_n $ is the radius of the circle in the second outer layer $\sin{\frac{\pi}{n}} = \frac{\hat{r}_n}{R_n -\hat{r}_n}$. The last two relations yield $R_2$ to be substituted in $(\star)$ whence we get an expression for $P_n$ and $$\lim_{n \to \infty} P_n = \frac{\pi}{2\sqrt{3}}$$
I agree with the author's answer in the first case. But in the second case, I got $$\lim_{n \to \infty} P_n =\displaystyle\frac{n \cdot r_n}{2R - r_n}$$ Now, how is this expression for $P_n$ equal to $\displaystyle\frac{\pi}{2\sqrt{3}}$
Is this final answer computed similar to that computed in this question titled Inscribed circles inside an equilateral triangle
If any member of Mathematics stack exchange knows satisfactory answers to my questions, may anwer them.