Find the area of the region enclosed between the two circles: x^2 + y…

The correct answer is Option (1) → $\frac{8\pi}{3} - 2\sqrt{3}$

Equations of the given circles are

$x^2 + y^2 = 4 \quad \dots (1)$

$\text{and} \quad (x - 2)^2 + y^2 = 4 \quad \dots (2)$

Equation $(1)$ is a circle with centre $O$ at the origin and radius $2$. Equation $(2)$ is a circle with centre $C(2, 0)$ and radius $2$. Solving equations $(1)$ and $(2)$, we have

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

$\text{or} \quad x^2 - 4x + 4 + y^2 = x^2 + y^2$

$\text{or} \quad x = 1 \text{ which gives } y = \pm \sqrt{3}$

Thus, the points of intersection of the given circles are $A(1, \sqrt{3})$ and $A'(1, -\sqrt{3})$ as shown in the Fig.

Required area of the enclosed region $OACA'O$ between circles:

$= 2 [\text{area of the region } ODCAO]$

$= 2 [\text{area of the region } ODAO + \text{area of the region } DCAD] \text{}$

$= 2 \left[ \int\limits_{0}^{1} y \, dx + \int\limits_{1}^{2} y \, dx \right] \text{}$

$= 2 \left[ \int\limits_{0}^{1} \sqrt{4 - (x - 2)^2} \, dx + \int\limits_{1}^{2} \sqrt{4 - x^2} \, dx \right]$

$= 2 \left[ \frac{1}{2} (x - 2) \sqrt{4 - (x - 2)^2} + \frac{1}{2} \times 4 \sin^{-1} \left( \frac{x - 2}{2} \right) \right]_{0}^{1} + 2 \left[ \frac{1}{2} x \sqrt{4 - x^2} + \frac{1}{2} \times 4 \sin^{-1} \frac{x}{2} \right]_{1}^{2} \text{}$

$= \left[ (x - 2) \sqrt{4 - (x - 2)^2} + 4 \sin^{-1} \left( \frac{x - 2}{2} \right) \right]_{0}^{1} + \left[ x \sqrt{4 - x^2} + 4 \sin^{-1} \frac{x}{2} \right]_{1}^{2} \text{}$

$= \left[ \left( -\sqrt{3} + 4 \sin^{-1} \left( \frac{-1}{2} \right) \right) - 4 \sin^{-1} (-1) \right] + \left[ 4 \sin^{-1} 1 - \sqrt{3} - 4 \sin^{-1} \frac{1}{2} \right] \text{}$

$= \left[ \left( -\sqrt{3} - 4 \times \frac{\pi}{6} \right) + 4 \times \frac{\pi}{2} \right] + \left[ 4 \times \frac{\pi}{2} - \sqrt{3} - 4 \times \frac{\pi}{6} \right] \text{}$

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

$= \frac{8\pi}{3} - 2\sqrt{3} \text{}$