Best way to split up a circle into subsections I'm not really sure how to word this question so feel free to edit it to make more sense:
I have a big circle of radius 32, and I want to divide it into smaller segments by both drawing smaller circle(s) with the same midpoint and drawing sectors (?) in such a way that each segment will fit into a circle of radius 9.5
I brought pictures!:
Imagine that if I were to divide the big circle into 8 slices and make 2 sub-circles of 20 and 10 radii, then you can see that it wont fit (the biege circle is the 9.5 radius one):


But if I make 16 slices, it will fit and I'll only need one smaller circle with a raidus of 18:


So my question is: is there any optimal way to figure out the least amount of cuts and sub-circles that are needed to make this work?
The cuts don't need to go all the way through the from the edge of the big circle to the center, it can start or stop at any smaller circle.
By the way, I used Blender to draw the pictures so it may not be exact.
 A: Let $R$ (=32) be the radius of the large circle and $r$ (=9.5) be the radius of the small disk.
We start with the following simpler problem: Given an annular sector $S$ with inner radius $a$, outer radius $b$ and central angle $\phi$, what is the radius $\rho$ of the smallest circle that covers $S$?

Using the above figure one gets 
$$\rho^2={a^2+b^2-2 a b\cos\phi\over2(1+\cos\phi)}\qquad(0<\phi<\pi)\ .$$
As suggested by Ross Millikan we now aim at minimizing the number of pieces. To this end I propose the following greedy algorithm:


*

*Put $r_0:=R$.

*Given $r_k$ for a $k\geq0$ put $b=r_k$, choose a suitable $n$ and put $\phi:={2\pi\over n}$. Then solve the equation
$$r^2={a^2+b^2-2 a b\cos\phi\over2(1+\cos\phi)}$$
for $a$ (choose the smaller of the two solutions).

*Repeat step 2. for various values of $n$ until you have found the $n$ maximizing  the quantity
$$q_n:={b^2-a^2\over n}\ .$$
(Comment: We want to maximize the area gain per piece.)

*If $a<1$ goto 5., else put $r_{k+1}:=a$ and goto 2.

*Cover the center with a last disk.

