Nine Circles

Since the system is symmetrical about the center line, let us consider half of the circles on the right. Let the radius of red circle be $r$ and the radius of the large yellow circle be $R$ . Consider the line $AE$ whose length is $R$ . Then

$\begin{aligned} AD + DE & = AE \\ AD + \sqrt{DG^2-GE^2} & = AE \\ 3r + \sqrt{(R+r)^2-(R-r)^2} & = R \\ 3r + 2\sqrt{rR} - R & = 0 \\ (3\sqrt r - \sqrt R)(\sqrt r + \sqrt R) & = 0 & \small \blue{\text{Since }R > r > 0} \\ 3 \sqrt r & = \sqrt R \\ \implies R & = 9 r \end{aligned}$

Also

$\begin{aligned} AC + CE & = AE \\ AC + FH & = AE \\ AC + \sqrt{FG^2 - GH^2} & = AE \\ AC + \sqrt{FG^2-(GI-HE-EI)^2} & = AE \\ 2r + \sqrt{(R+10)^2 - (R - FC - r)^2} & = R \\ 2r + \sqrt{(9r+10)^2 - (8r - \sqrt{(r+10)^2-r^2})^2} & = 9r \\ \sqrt{(9r+10)^2 - (8r - \sqrt{20r+100})^2} & = 7r \\ 81r^2 + 180r + 100 - (64r^2 - 16r\sqrt{20r+100} + 20r + 100) & = 49r^2 \\ 16r\sqrt{20r+100} & = 32r^2 - 160r & \small \blue{\text{Since }r > 0} \\ \sqrt{5r+25} & = r - 5 & \small \blue{\text{Squaring both sides.}} \\ 5r + 25 & = r^2 - 10r + 25 \\ \implies r & = 15 \\ R & = 9r = \boxed{135} \end{aligned}$