circleseuclidean-geometry
Here is the picture of what I meant (the small circles are all the same radius, I'm just really bad at paint)
So I realised that you can connect the circle centers to make a hexagon with all sides equal to $2r$ where $r$ is the radius of the small circle, but in order to prove that the radius of the big circle is 3 times larger than $r$, i have to prove that the hexagon is regular, but I don't know how to prove that the angles of the hexagon are all equal.
Let $C_1$ be the center of the circle you want (where you only know its $x$-coordinate), $r_1$ be the radius of that circle, and $C_2$, $r_2$, $C_3$, and $r_3$ be the center and radius of the second and third circles. It appears that the circle you want will be internally tangent to each of the other circles, so the distance between $C_1$ and $C_2$ must be $r_2-r_1$ and the distance between $C_1$ and $C_3$ must be $r_3-r_1$. Solving the system generated by these two requirements gives: $$r_1=0.763622\quad\text{ and }\quad C_1=(2.027134,5.80275)$$ or $$r_1=0.763622\quad\text{ and }\quad C_1=(2.027134,8.08365)$$ (solved using Mathematica).
edit If the circle we're finding is externally tangent to the second circle and internally tangent to the third circle, then the distance between $C_1$ and $C_2$ must be $r_1+r_2$ and the distance between $C_1$ and $C_3$ must be $r_3-r_1$. Solving the system generated by these two requirements gives: $$r_1=0.410104\quad\text{ and }\quad C_1=(2.027134,4.62903)$$ or $$r_1=0.410104\quad\text{ and }\quad C_1=(2.027134,9.25737)$$ (solved using Mathematica).
Here's a picture of the case $n=5$. 
If $R$ is the radius of the big circle and $r$ the radii of the small circles, do you see that the isosceles triangle has two sides $R-r$ and one side $2r$?
Best Answer
By symmetry, the triangle formed by the centers of three tangent circles is equilateral. Then in the figure formed by seven circles, the alignments are perfect ($3\times60°=180°$) and the large diameter is three times a small one.