Shad-ircles

$\textcolor{#EC7300}{\text{Orange Areas}}$

$\begin{aligned} \textcolor{#EC7300}{\text{Sum of Orange Areas}} &= (\pi8^2 - \pi7^2) + (\pi6^2 - \pi5^2) + (\pi4^2 - \pi3^2) + (\pi2^2 - \pi1^2) \\ &= \pi(8^2 - 7^2 + 6^2 - 5^2 + 4^2 - 3^2 + 2^2 - 1^2) \\ &= \pi(36) \end{aligned}$

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$\textcolor{#3D99F6}{\text{Blue Areas}}$

$\begin{aligned} \textcolor{#3D99F6}{\text{Sum of Blue Areas}} &= (\pi7^2 - \pi6^2) + (\pi5^2 - \pi4^2) + (\pi3^2 - \pi2^2) + (\pi1^2) \\ &= \pi(7^2 - 6^2 + 5^2 - 4^2 + 3^2 - 2^2 + 1^2) \\ &= \pi(28) \end{aligned}$

$\text{OR}$

$\begin{aligned} \textcolor{#3D99F6}{\text{Sum of Blue Areas}} &= \text{Area of Bigger circle} - \textcolor{#EC7300}{\text{Orange Areas}} \\ &= \pi(64) - \pi(36) \\ &= \pi(28) \end{aligned}$

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$\therefore \frac{\textcolor{#EC7300}{\text{Orange Areas}}}{\textcolor{#3D99F6}{\text{Blue Areas}}} = \frac{\pi(36)}{\pi(28)} = \boxed{\frac{9}{7}}$

Let the areas of blue and orange regions be $A_b$ and $A_o$ respectively. Then we have:

$\begin{aligned} A_b & = \pi \left(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + 7^2 \right) \\ & = \pi \left( \sum_{n=1}^7 n^2 - 2\sum_{n=1}^3 (2n)^2 \right) \\ & = \pi \left(\frac {7(8)(15)}6 - 2 \times 4 \times \frac {3(4)(7)}6 \right) \\ & = 28 \pi \end{aligned}$

$\begin{aligned} A_o & = \pi \left(\blue{- 1^2 + 2^2 - 3^2 + 4^2 - 5^2 + 6^2 - 7^2} + 8^2 \right) \\ & = \blue{- A_b} + 8^2 \pi = -28 \pi + 64 \pi = 36 \pi \end{aligned}$

Therefore, $A_o : A_b = 36 \pi : 28 \pi = \boxed{9:7}$ .