2010 unit circles $C$ is a unit circle with radius $r$. $C_1,C_2,\ldots, C_{2010}$ are unit circles along the circumference of $C$ touching $C$ externally. Also the pairs $C_1C_2;C_2C_3,\ldots;C_{2010}C_1$ touch. Then find $r$. Options:
1) $cosec(\pi/2010)$
2) $sec(\pi/2010)$
3) $cosec(\pi/2010)-1$
4) $sec(\pi/2010)-1$
I try solving but I am getting a weird angle. A picture might help. Thank you.
 A: 
Made it before I noticed its already done, but anyway here it is...
A: OK, here's a picture.   The right triangle has hypotenuse $\ldots$, one angle $\ldots$, and the opposite side $\ldots$. 

A: 
The generalized formula for n. no of circles each with a radius $R$ & touching one another along the circumference of a circle with a radius $r$ (above diagram is only for visualization) is given as follows
$$\bbox[4pt, border: 1px solid blue;]{r=R\left[cosec\frac{\pi}{n}-1\right]}$$
where, $n$ is a natural number $n\geq3$
Above formula can be easily derived by joining the centers $C_1, C_2, C_3, \!  \cdots \! ....,C_{n-1}, C_n$ of all the circles & analyzing the polygon obtained 
Now, for 2010 unit circles, substitute $n=2010$ & $R=1$ in the above generalized formula, we can easily find out the radius $r$ of circle C as follows
$$r=R\left[cosec\frac{\pi}{n}-1\right]=(1)\left[cosec\frac{\pi}{2010}-1\right]=cosec\frac{\pi}{2010}-1$$
Hence the option (3) is correct.
A: Educated guess:
The $2010$ centers are uniformly spread on a circle of radius $r+1$, so there will most probably be a $-1$ term. And the trigonometric function must be a $\text{cosec}$ as it needs to go to $\infty$ as $2010$ is increased.
My bet for $3$.
