Area Under the Curve
Area Between Two Circles
nta_pyq_2026_jan
Grade 12

Question:

The area of the region enclosed between the circles $x^2+y^2=4$ and $x^2+(y-2)^2=4$ is:
$\dfrac{4}{3}(2\pi-3\sqrt{3})$
$\dfrac{2}{3}(2\pi-3\sqrt{3})$
$\dfrac{2}{3}(4\pi-3\sqrt{3})$
$\dfrac{4}{3}(2\pi-\sqrt{3})$

Step-by-Step Solution

Key Concept: Circle 1: centre $(0,0)$, $r=2$. Circle 2: centre $(0,2)$, $r=2$. Distance between centres $=2$. Intersection at $y=1$, $x=\pm\sqrt{3}$. Angle subtended at centre of circle 1: $\theta=2\pi/3$.
Intersection area $=\frac{2}{3}(4\pi-3\sqrt{3})$.
Correct Answer: 3

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