Family Of Circles

Not counting the circles of radius $3$ and $2$ , let $k_0,k_1,k_2,\ldots,k_4$ be the curvatures of the other circles. from largest to smallest. Then $k_0 = 1$ . For any $0 \le n \le 3$ , we see that the circle of curvature $\tfrac13$ is externally tangent to the three mutually tangent circles of curvature $\tfrac12$ , $k_n$ and $k_{n+1}$ . Using Descartes Theorem, this implies that $-\tfrac13 \; = \; k_{n+1} + k_n + \tfrac12 - 2\sqrt{k_{n+1}k_n + \tfrac12k_n + \tfrac12k_{n+1}}$ and hence that $k_{n+1} \; = \; \tfrac16 + k_n + \tfrac13\sqrt{6(k_n - 1)}$ A simple induction shows that $k_n \; = \; \tfrac16(6 + n^2) \hspace{1cm} 0 \le n \le 4$ and hence $k_4 = \tfrac{11}{3}$ , so the radius of the smallest circle is $\tfrac{3}{11} = \boxed{0.\dot{2}\dot{7}}$ .

• If the radius BY=a, AX =b
• The radius of larger circle is (a+b)
• Leaving these 3 circles, there is a generalised formula for calculating the radius of any circle tangent to these circles and the pattern continuing! The radius of nth circle is--
• $(a\times b\times (a+b))\div (n^{2}\times a^{2} + a\times b + b^{2} )$
• so accordingly, radius of 4th circle in the pattern is-
• $(1\times 2\times 3)/(16\times 1 + 2 + 4 )$ = 0.272
• $\textbf{0.27}$

$Using~ Descartes ~Theorem,\\ K_1= - 1/3,~~~K_2= 1/2,~~~K_3= 1/1.\\ \therefore~K_4 = - 1/3 + 1/2 + 1/1 \pm ~2\sqrt{- 1/6+1/2 - 1/3}.\\ Gives~\color{#3D99F6}{K_4=7/6}.\\ For~next~circles,~K_3~ will ~be~replaced~by~K_n,~and ~K_4 ~by~K_{n+1},~~~n=4,~5, ~6,~ 7.~\\ So~K_{n+1}=- 1/3 + 1/2 + K_n \pm ~2\sqrt{- 1/6+1/2 *K_n - 1/3*K_n.}\\ \implies~\color{#3D99F6}{K_{n+1}= 1/6 + K_n \pm ~2\sqrt{\dfrac{K_n - 1} 6} }. \\ n=4,~~K_5=1/6 + 7/6 \pm~\sqrt{\dfrac{7/6-1} 6}=4/3\pm 1/3. ......~~K_5=5/3,~~~ K_3=1/1..(correct) \\ \color{#3D99F6}{K_5=5/3}.\\ n=5,~~K_6=1/6 + 5/3 \pm~\sqrt{\dfrac{5/3-1} 6}=11/6 \pm 2/3...... ~~K_6=5/2, ~~~K_4=7/6..(correct) \\ \color{#3D99F6}{K_6=5/2}.\\ n=6,~~K_7=1/6 + 5/2 \pm~\sqrt{\dfrac{5/2-1} 6}=8/3 \pm 1. ~~.....K_7=11/3, ~~~K_4=5/3..(correct) \\ \color{#D61F06}{K_7=11/3}.\\ R_7=\Large~~\color{#D61F06}{0.272727}.$

Note that at every $n^{th}$ circle we get two values. The - tive gives value of $(n-1)^{th}$ circle, and that confirms with our previous value, tells us that our calculations are correct.