JEE Mathematics 2026 Question with Solution

Given: The circles are $x^2 + y^2 = 4$ and $x^2 + (y - 2)^2 = 4$.

Find: The area of the region enclosed between the two circles.

The circles intersect where

$$x^2+y^2 = x^2+(y-2)^2$$

So,

$$y^2 = y^2 - 4y + 4$$ $$4y = 4$$ $$y = 1$$

Hence, the intersection points are $\left(\pm\sqrt{3}, 1\right)$.

Consider the sector of the circle $x^2+y^2=4$ cut by the chord $y=1$. Its center is $\left(0,0\right)$ and radius is $2$. The distance of the chord from the center is $1$.

Let the half-angle be $\alpha$. Then

$$\cos \alpha = \frac{1}{2}$$

so

$$\alpha = \frac{\pi}{3}$$

Therefore, the total angle subtended is

$$2\alpha = \frac{2\pi}{3}$$

Area of one circular segment is

$$A_{seg} = \frac{1}{2}r^2(2\alpha) - \frac{1}{2}r^2\sin(2\alpha)$$

Substituting $r=2$ and $2\alpha = \frac{2\pi}{3}$,

$$A_{seg} = \frac{1}{2}(4)\left(\frac{2\pi}{3}\right) - \frac{1}{2}(4)\left(\frac{\sqrt{3}}{2}\right)$$ $$A_{seg} = \frac{4\pi}{3} - \sqrt{3}$$

The total enclosed area is made of two identical segments, one from each circle. Hence,

$$\text{Total Area} = 2\left(\frac{4\pi}{3} - \sqrt{3}\right)$$ $$\text{Total Area} = \frac{8\pi}{3} - 2\sqrt{3}$$

Factoring out $\frac{2}{3}$,

$$\text{Total Area} = \frac{2}{3}(4\pi - 3\sqrt{3})$$

Therefore, the correct option is B.

Given: Two equal circles $x^2 + y^2 = 4$ and $x^2 + (y - 2)^2 = 4$ intersect symmetrically.

Find: The common enclosed area.

Because the circles have the same radius and their centers are symmetrically placed, the common region consists of two identical circular segments.

First find the intersection line by equating the two circle equations:

$$y = 1$$

So the common chord is at distance $1$ from each center. With radius $2$,

$$\cos \alpha = \frac{1}{2} \Rightarrow \alpha = \frac{\pi}{3}$$

Hence, the segment angle is

$$2\alpha = \frac{2\pi}{3}$$

Now compute one segment directly as sector minus triangle:

$$\frac{1}{2}(2)^2\left(\frac{2\pi}{3}\right) - \frac{1}{2}(2)^2\sin\left(\frac{2\pi}{3}\right)$$ $$= \frac{4\pi}{3} - \sqrt{3}$$

Doubling it gives

$$2\left(\frac{4\pi}{3} - \sqrt{3}\right) = \frac{2}{3}(4\pi - 3\sqrt{3})$$

Therefore, the correct option is B.