Let A = {(x, y) in R^2 : y geq 0, 2x leq y le | vedclass

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(A) The circle is given by $(x-1)^2 + y^2 = 4$,which has center $(1, 0)$ and radius $r = 2$.For set $A$,the region is bounded by $y = 2x$ and the uppe

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$\frac{\pi-1}{\pi+1}$

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(A) The circle is given by $(x-1)^2 + y^2 = 4$,which has center $(1, 0)$ and radius $r = 2$.For set $A$,the region is bounded by $y = 2x$ and the upper arc of the circle. The intersection of $y = 2x$ and $(x-1)^2 + y^2 = 4$ is found by substituting $y=2x$: $(x-1)^2 + 4x^2 = 4 \implies x^2 - 2x + 1 + 4x^2 = 4 \implies 5x^2 - 2x - 3 = 0 \implies (5x+3)(x-1) = 0$. Since $y \geq 0$,we take $x=1$,which gives $y=2$. The point of intersection is $(1, 2)$.The area of $A$ is the area under the circular arc from $x=0$ to $x=1$ minus the area of the triangle formed by $(0,0), (1,0), (1,2)$.Area of $A = \int_{0}^{1} \sqrt{4-(x-1)^2} dx - 	ext{Area}(	riangle OAB) = \frac{1}{4}(\pi 	imes 2^2) - \frac{1}{2}(1)(2) = \pi - 1$.For set $B$,the region is bounded by $y = 2x$ for $x \in [0, 1]$ and the circular arc for $x > 1$. The area is the sum of the area of the triangle with vertices $(0,0), (1,0), (1,2)$ and the area under the circular arc from $x=1$ to $x=3$.Area of $B = \frac{1}{2}(1)(2) + \int_{1}^{3} \sqrt{4-(x-1)^2} dx = 1 + \frac{1}{4}(\pi 	imes 2^2) = 1 + \pi$.The ratio of the area of $A$ to the area of $B$ is $\frac{\pi-1}{\pi+1}$.

Let $A = \{(x, y) \in R^2 : y \geq 0, 2x \leq y \leq \sqrt{4-(x-1)^2}\}$ and $B = \{(x, y) \in R \times R : 0 \leq y \leq \min \{2x, \sqrt{4-(x-1)^2}\}\}$. Then the ratio of the area of $A$ to the area of $B$ is

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