I would like to it is in able to readjust the variety of smaller circles and also the radius that the bigger circle.

As an example:

$$eginalignR&=1.5\n&=9\r&=,?endalign$$

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If you attract $n$ lines native the beginning touching the little circles and also $n$ lines indigenous the beginning to the center of each little circle friend basically divide the $2 pi$ angle into $2n$ same angles, to speak $ heta$. Hence $ heta=pi/n$. Currently a triangle v vertices the origin, the center of among the tiny circles and also a tangent point of the very same circle is a best triangle since the tangent is perpendicular come the radius in ~ the suggest of contact. You climate have

$$sin heta=fracrR$$

Putting the all together

$$r=R sin fracpin$$

Another approach: lets say we have $n$ tiny circles. Then the facility points that the little circles kind a constant $n$-gon, whereby the side size is $2r$. The radius of the huge circle is the circumradius of the $n$-gon i beg your pardon is $R = frac2r2sin(pi/n)$ so $r = R sin(pi/n)$

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Suppose the the center $c_k,k=1,ldots,n$ of the little circles are inserted equidistantly ~ above the bigger circle.

Then we have $c_k=left(Rsin(2pifrackn),Rcos(2pifrackn) ight), k = 1,ldots,n$. So because that $k=1,ldots,n$ we must have actually $r=frac2$, due to the fact that two one with same radius room tangent if your radius is the fifty percent of the distance between their centers. In details $$r=frac2=frac12sqrtig(Rsin(2pi)-Rsinig(2pifracn-1nig)ig)^2+ig(Rcos(2pi)-Rcosig(2pifracn-1nig)ig)^2 \ =fracR2sqrtsinig(2pifracn-1nig)^2+ig(1-cosig(2pifracn-1nig)ig)^2 \ = fracR2sqrt2-2cosig(2pifracn-1nig) =fracR2sqrt2Big(1-cosig(2pifracn-1nig)Big)\=fracR2sqrt4sinig(2pifracn-1nig)^2=Rsinig(pifracn-1nig)=Rsinig(fracpinig).$$