Animated Solution for Mathematics - Definite Integration: The area of the region enclosed between the circles $x^2+y^2=4$ and $x^2+(y-2{)}^2=4$ is:

Welcome, fellow traveler of the JEE path. Today, we are not just solving a problem; we are exploring the elegant dance of two circles.

Imagine standing on a coordinate plane. You have one circle, $x^2+y^2=4$, sitting comfortably at the origin. It is a perfect, symmetric entity with a radius of $r=2$.

Now, imagine a second circle, $x^2+(y-2{)}^2=4$, identical in size but shifted upward so its center rests at $(0,2)$. Where they overlap, they create a beautiful, lens-shaped region. This is our target.

To begin our journey, we must find where these two worlds collide. We set the equations equal:

Since both equal $4$, we can write $x^2+y^2=x^2+(y-2{)}^2$. The $x^2$ terms vanish, a moment of algebraic grace that simplifies our path.

We are left with $y^2=(y-2{)}^2$. Expanding the right side, we get $y^2=y^2-4y+4$. The $y^2$ terms cancel, leaving $4y=4$, or simply $y=1$.

This horizontal line, $y=1$, is the common chord. It is the backbone of our region. Substituting $y=1$ back into our first circle, we find $x^2+1=4$, so $x^2=3$, giving us $x=\pm \sqrt{3}$.

Our intersection points are $(\sqrt{3},1)$ and $(-\sqrt{3},1)$. We have mapped the territory.

Now, we face the region. Because the circles are identical, the area above the line $y=1$ is a perfect mirror of the area below it.

We only need to calculate the area of the upper segment of the first circle and multiply it by $2$. This is the 'JEE mindset'—finding the most efficient, elegant path through the forest.

We focus on the upper segment of the circle $x^2+y^2=4$. We draw radii from the center $(0,0)$ to the intersection points $(\sqrt{3},1)$ and $(-\sqrt{3},1)$. These radii have length $r=2$.

The distance from the center to the chord $y=1$ is $h=1$. We have a triangle with sides $2,2$ and a base of $2\sqrt{3}$.

The central angle $\theta$ is what we need. Using the right triangle formed by the perpendicular, we see:

Thus, $\frac{\theta }{2}=\frac{\pi }{3}$, which means $\theta =\frac{2\pi }{3}$.

We are at the finish line. The area of a circular segment is the area of the sector minus the area of the triangle:

Substituting our values, $r=2$ and $\theta =\frac{2\pi }{3}$, we get:

Since $\sin \left(\frac{2\pi }{3}\right)=\frac{\sqrt{3}}{2}$, this becomes:

Finally, we multiply by $2$ to account for both segments:

Factoring out $\frac{2}{3}$, we arrive at the final result:

You have navigated the geometry, mastered the algebra, and utilized symmetry to conquer the problem. Take a moment to appreciate the result. It is not just a number; it is the measure of the space between two circles, perfectly captured.