1. Interpret the regions

We are given $A=\{(x,y)\in\mathbb R^2: y\ge 0,\ 2x\le y\le \sqrt{4-(x-1)^2}\}$ and $B=\{(x,y)\in\mathbb R^2: 0\le y\le \min\{2x,\sqrt{4-(x-1)^2}\}\}.$

The curve $y=\sqrt{4-(x-1)^2}$ is the upper semicircle of the circle $(x-1)^2+y^2=4,$ which has center $(1,0)$ and radius $2$.

The line is $y=2x.$

Both regions lie above the $x$-axis.

2. Find intersection points of the line and semicircle

Set $2x=\sqrt{4-(x-1)^2}.$ Since the square root is nonnegative, we must have $x\ge 0$.

Squaring: $4x^2=4-(x-1)^2.$ Now, $(x-1)^2=x^2-2x+1,$ so $4x^2=4-(x^2-2x+1)=3-x^2+2x.$ Thus, $5x^2-2x-3=0.$ Solving: $x=\frac{2\pm\sqrt{4+60}}{10}=\frac{2\pm 8}{10}.$ Hence, $x=1\quad \text{or} \quad x=-\frac35.$ But $x=-\frac35$ is invalid because then $2x<0$ while the square root is nonnegative. So the relevant intersection is $(1,2).$

3. Understand region $B$

Region $B$ is the area under the smaller of the two curves: $y=2x \quad \text{and} \quad y=\sqrt{4-(x-1)^2},$ with $y\ge 0$.

The semicircle exists for $-1\le x\le 3.$ For $x<0$, we have $2x<0$, so $\min\{2x,\sqrt{4-(x-1)^2}\}<0$, and the condition $0\le y\le \min\{\cdots\}$ is impossible. Hence $B$ only exists for $x\ge 0$.

From the intersection analysis:

• for $0\le x\le 1$, we have $2x\le \sqrt{4-(x-1)^2}$,
• for $1\le x\le 3$, we have $\sqrt{4-(x-1)^2}\le 2x$.

Therefore, $\text{Area}(B)=\int_0^1 2x\,dx+\int_1^3 \sqrt{4-(x-1)^2}\,dx.$

Compute: $\int_0^1 2x\,dx=[x^2]_0^1=1.$

For the second integral, let $u=x-1$. Then when $x=1$, $u=0$, and when $x=3$, $u=2$: $\int_1^3 \sqrt{4-(x-1)^2}\,dx=\int_0^2 \sqrt{4-u^2}\,du.$ This is the area of a quarter circle of radius $2$: $\int_0^2 \sqrt{4-u^2}\,du=\frac{\pi (2)^2}{4}=\pi.$ So, $\boxed{\text{Area}(B)=1+\pi.}$

4. Understand region $A$

Region $A$ satisfies $2x\le y\le \sqrt{4-(x-1)^2}, \quad y\ge 0.$ So it is the part between the semicircle and the line where the semicircle lies above the line.

This occurs for $x\in[-1,1]$, but we must also respect $y\ge 0$ and $y\ge 2x$.

A simpler observation is:

• The entire upper semicircle area is $\frac12\pi(2)^2=2\pi.$
• This upper semicircle is partitioned into two disjoint parts: region $A$ and region $B$.

Indeed, for each $x$ where the semicircle exists and $y\ge 0$, the vertical segment from $y=0$ to the semicircle is split by the lower boundary $y=2x$ into:

• the part below the minimum curve = region $B$,
• the part above the line and below semicircle = region $A$.

Hence, $\text{Area}(A)+\text{Area}(B)=2\pi.$ Therefore, $\text{Area}(A)=2\pi-(\pi+1)=\pi-1.$

5. Compute the required ratio

$\frac{\text{Area}(A)}{\text{Area}(B)}=\frac{\pi-1}{\pi+1}.$

So the correct option is $\boxed{\text{B }\frac{\pi-1}{\pi+1}}.$

6. Comparison with stored answer

Stored correct answer: B

Our derived answer is also B. Hence they agree.