Circle Geometry Question

1) Double theta so that the side opposite it connects the centers of adjacent circles. There will be n such segments.
2*theta = 360/n
theta = 180/n

2) The triangle in the sketch is right because the radius (side r) is perpendicular to the tangent.
sin(theta) = r/(R+r)
R*sin(theta) + r*sin(theta) = r
R*sin(theta) = r*(1 - sin(theta))
R/r = (1 - sin(theta))/sin(theta)

3) Since the triangle is right,
r^2 + (R+x)^2 = (R+r)^2
x^2 + 2Rx - 2Rr = 0
x = sqrt(R^2 + 2Rr) - R

4) From (3) you have an expression for x in terms of R and r. From (2) you have an expression for r in terms of R and theta. From (1) you have an expression for theta in terms of n. Combine them, and plug 4, 6, and 12 into the resulting expression.

5) You want r<R. From (2) sin(theta) < 1 - sin(theta). Figure out how large theta can be (larger theta implies fewer surrounding circles). From (1) you can compute n.

I come up with the same solution as markr. But to elaborate
on 4) I'd substitute
R=r*sin(180/n)/[1-sin(180/n)]
into
x=sqrt(R^2+2rR)-2R
to get x in terms of R and n
x=R{sqrt[(1+sin(180/n))/(1-sin(180/n))]-2}
& 5)
set r=R and solve for n
or 1=[1-sin(180/n)]/ sin(180/n)
sin(180/n)=1-sin(180/n)
sin(180/n)=1/2
180/n-asin(1/2)=30
n=180/30=6
do for outer circles to be the same size as the inner n=6. If the outer circles are smaller n>6.

Rap

Thanks alot for your help. Could you prehaps just elaborate on question 4? I see your reasoning here but putiing it in question form 4.1 -> 4.3

Would you just substitute the values 4,6 and 12 into:

x = R(sqrt)[(1+sin(180/n))/(1-sin(180/n))]-2}

plug in 4, 6, and 12 for n into
x=R{sqrt[(1+sin(180/n))/(1-sin(180/n))]-2}

Rap

quick question, wouldn't 6 ensure that they're all the exact same size as the center one? wouldn't you need seven or more to ensure that they're smaller?

Yep. And Rap's calculations confirm it.

Yeah. n=6 therefor n>7 so 7 is the maximum value I believe.

Thanks for all the input anyway.