Animated Solution for Mathematics - Definite Integration: Let $A=\{(x,y)\in {\mathbb{R}}^2:y\ge 0,2x\le y\le \sqrt{4-(x-1{)}^2}\}$ and $B=\{(x,y)\in \mathbb{R}\times \mathbb{R}:0\le y\le \min \{2x,\sqrt{4-(x-1{)}^2}\}\}$. Then the ratio of the area of $A$ to the area of $B$ is

Imagine you are standing on the coordinate plane, looking at two distinct curves that define a beautiful, hidden landscape. We have a circle,

This is a perfect, symmetric shape centered at $(1,0)$ with a radius of $2$. Cutting through this landscape is a straight line, $y=2x$, originating from the origin.

This isn't just a math problem; it's a test of your ability to see the simplicity hidden within complexity. Many students would immediately reach for the integral sign, but let's pause and look at the geometry.

First, we must find where these two paths meet. By substituting $y=2x$ into the circle's equation, we get:

Expanding this, we find $x^2-2x+1+4x^2=4$, which simplifies to the quadratic:

Factoring this, we get $(x-1)(5x+3)=0$. Since our region is defined for $y\ge 0$, we focus on the positive intersection at $x=1$, which gives us the point $(1,2)$. This point is our anchor; it is the key to unlocking the entire geometry of the problem.

Now, let's define Region A. It is the area where $2x\le y\le \sqrt{4-(x-1{)}^2}$. Graphically, this is the space trapped between the line and the upper arc of the semi-circle.

Instead of integrating, look at the shapes. We have a left quarter circle spanning from $x=-1$ to $x=1$. If we take this quarter circle and subtract the area of the triangle formed by the line $y=2x$ from $x=0$ to $x=1$, we are left with exactly Region A.

The area of the quarter circle is:

The area of the triangle is $\frac{1}{2}\times 1\times 2=1$. Thus, the area of Region A is simply $\pi -1$.

Next, we turn our attention to Region B, defined by $0\le y\le \min \{2x,\sqrt{4-(x-1{)}^2}\}$. This region is the area below the lower of the two curves.

From $x=0$ to $x=1$, the line is lower, forming that same triangle with area $1$. From $x=1$ to $x=3$, the semi-circle is lower, forming the right quarter circle with area $\pi$.

Adding these two parts together, the total area of Region B is $1+\pi$.

We have arrived at the final step. We need the ratio of the area of Region A to the area of Region B. With our calculated values, this is:

Look at that result. It is a testament to the power of geometric intuition. By avoiding the trap of brute-force integration, we found the answer with clarity and confidence.

Remember, in JEE Advanced, the most powerful tool in your arsenal is not just your ability to calculate, but your ability to visualize the geometry behind the equations. Keep practicing, keep visualizing, and you will master these problems with ease.