i am trying come determine how to calculate the required radius that the smaller sized circles therefore they touch every other around the larger circle. (red box)

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\$\$  You are watching: Smaller circles in a larger circle  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)\$

See more: How To Download And Burn Xbox 360 Games On To A Blank Disc, How To Properly Burn Xbox 360 Games 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=frac12sqrtig(Rsin(2pi)-Rsinig(2pifracn-1nig)ig)^2+ig(Rcos(2pi)-Rcosig(2pifracn-1nig)ig)^2 \ =fracR2sqrtsinig(2pifracn-1nig)^2+ig(1-cosig(2pifracn-1nig)ig)^2 \ = fracR2sqrt2-2cosig(2pifracn-1nig) =fracR2sqrt2Big(1-cosig(2pifracn-1nig)Big)\=fracR2sqrt4sinig(2pifracn-1nig)^2=Rsinig(pifracn-1nig)=Rsinig(fracpinig).\$\$