2006 AMC 10B Problem 4

1. Find each area. The red circle has diameter $1$, so radius $\frac12$ and area $\pi \cdot \left(\frac12\right)^2 = \frac{\pi}{4}$.
2. The large circle has diameter $3$, so radius $\frac32$ and area $\pi \cdot \left(\frac32\right)^2 = \frac{9\pi}{4}$.
3. The blue part is the large circle with the red circle cut out of its middle. So
$$\text{blue} = \frac{9\pi}{4} - \frac{\pi}{4} = \frac{8\pi}{4}.$$
4. Divide blue by red:
$$\frac{8\pi/4}{\pi/4} = 8.$$

The answer is $\boxed{\textbf{(D)}\ 8}$.

When one circle is $3$ times as wide as another, its area is $3 \times 3 = 9$ times as big, because area grows with the square of the width. So you never really need the radii: the big circle is $9$ "red circles" of area, and removing the one red circle in the middle leaves $8$. Look for ring-shaped regions as "big minus small."

The diameters are in ratio $3 : 1$, so the areas of the two full circles are in ratio $3^2 : 1^2 = 9 : 1$. Call the red area $R$; then the large circle has area $9R$.

The blue region is everything inside the large circle but outside the small one, so its area is $9R - R = 8R$.

The ratio of blue to red is $\frac{8R}{R} = 8$.

The answer is $\boxed{\textbf{(D)}\ 8}$.