### Instruction: An arc of $55$ degrees on circle $A$ has the same length as an arc of $40$ degrees on circle $B$.  What is the ratio of the area of circle $A$ to the area of circle $B$? Express your answer as a common fraction.### Response: To compare the areas of the circles, I need to find their radii.
I recall that the length of an arc is proportional to the central angle and the radius of the circle, so I can use the formula $s = r\theta$, where $s$ is the arc length, $r$ is the radius, and $\theta$ is the central angle in radians.
I also recall that one radian is equivalent to $180/\pi$ degrees, so I can convert the given angles to radians by multiplying by $\pi/180$.
Let $s_A$ and $s_B$ be the lengths of the arcs on circles $A$ and $B$, respectively, and let $r_A$ and $r_B$ be their radii.
Then I have the equations $s_A = r_A \cdot \frac{55\pi}{180}$ and $s_B = r_B \cdot \frac{40\pi}{180}$.
Since the problem says that the arc lengths are equal, I can set $s_A = s_B$ and get $r_A \cdot \frac{55\pi}{180} = r_B \cdot \frac{40\pi}{180}$.
I can simplify this equation by canceling out the common factors of $\pi$ and $180$, and get $r_A \cdot 55 = r_B \cdot 40$.
Solving for $r_A$, I get $r_A = \frac{40}{55} r_B = \frac{8}{11} r_B$.
Now I can find the ratio of the areas of the circles by using the formula $A = \pi r^2$.
The ratio of the areas is $\frac{A_A}{A_B} = \frac{\pi r_A^2}{\pi r_B^2} = \frac{r_A^2}{r_B^2}$.
Substituting the expression for $r_A$ in terms of $r_B$, I get $\frac{A_A}{A_B} = \frac{(\frac{8}{11} r_B)^2}{r_B^2} = \frac{64}{121} \cdot \frac{r_B^2}{r_B^2} = \frac{64}{121}$.
This is the final answer.