Q36 / 75 · JEE Main 2026 (23 January Shift 2) · Mathematics
The area of the region enclosed between the circles $x^{2}+y^{2}=4$ and $x^{2}+(y-2)^{2}=4$ is:
- A$\frac{4}{3}(2 \pi-3 \sqrt{3})$
- B$\frac{2}{3}(2 \pi-3 \sqrt{3})$
- Correct answerC$\frac{2}{3}(4 \pi-3 \sqrt{3})$
- D$\frac{4}{3}(2 \pi-\sqrt{3})$
Solution
Circle 1: centre $(0,0)$, $r=2$. Circle 2: centre $(0,2)$, $r=2$. Distance between centres $d=2$.
Intersection: $y^2 = (y-2)^2 \Rightarrow y = 1$, giving $x = \pm\sqrt{3}$.
Angle subtended at centre of circle 1 by the chord: $\theta = 2\pi/3$.
Segment area $= \frac{1}{2}(4)\left(\frac{2\pi}{3}\right) - \frac{1}{2}(4)\sin\frac{2\pi}{3} = \frac{4\pi}{3} - \sqrt{3}$.
By symmetry, total intersection area $= 2\left(\frac{4\pi}{3} - \sqrt{3}\right) = \frac{8\pi - 6\sqrt{3}}{3} = \frac{2}{3}(4\pi - 3\sqrt{3})$.
Intersection: $y^2 = (y-2)^2 \Rightarrow y = 1$, giving $x = \pm\sqrt{3}$.
Angle subtended at centre of circle 1 by the chord: $\theta = 2\pi/3$.
Segment area $= \frac{1}{2}(4)\left(\frac{2\pi}{3}\right) - \frac{1}{2}(4)\sin\frac{2\pi}{3} = \frac{4\pi}{3} - \sqrt{3}$.
By symmetry, total intersection area $= 2\left(\frac{4\pi}{3} - \sqrt{3}\right) = \frac{8\pi - 6\sqrt{3}}{3} = \frac{2}{3}(4\pi - 3\sqrt{3})$.
Hence, the correct answer is option: 'C'.