Question

Find the area of the region between the circles x² + y² = 4 and (x − 2)² + y² = 4.

Answer 1

\[x^2 + y^2 = 4\]   .......(1)       represents a circle with centre O(0,0) and radius  2

\[\left( x - 2 \right)^2 + y^2 = 4\] ......(2)   represents a circle with centre A(2 ,0) and radius 2

Points of intersection  of two circles is given by solving the equations
\[\left( x - 2 \right)^2 = x^2 \]
\[ \Rightarrow x^2 - 4x + 4 = x^2 \]
\[ \Rightarrow x = 1 \]
\[ \therefore y^2 = 3 \]
\[ \Rightarrow y = \pm \sqrt{3}\]
\[\text{ Now, B}\left( 1, \sqrt{3} \right)\text{ and B'}\left( 1, - \sqrt{3} \right)\text{ are two points of intersection of the two circles }\]
\[\text{ We need to find shaded area }= 2 \times\text{ area }\left(\text{ OBAO }\right) . . . \left( 1 \right)\]
\[\text{ Area }\left(\text{ OBAO }\right) = \text{ area }\left(\text{ OBPO }\right) +\text{ area }\left(\text{ PBAP }\right) \]
\[ = \int_0^1 \left| y_1 \right|dx + \int_1^2 \left| y_2 \right|dx ..............\left\{ \because y_1 > 0 \Rightarrow \left| y_1 \right| = y_1\text{ and }y_2 > 0 \Rightarrow \left| y_2 \right| = y_2 \right\}\]
\[ = \int_0^1 y_1 dx + \int_1^2 y_2 dx\]
\[ = \int_0^1 \sqrt{4 - \left( x - 2 \right)^2} dx + \int_1^2 \sqrt{4 - x^2} dx\]
\[ = \left[ \frac{1}{2}\left( x - 2 \right)\sqrt{4 - \left( x - 2 \right)^2} + \frac{1}{2} \times 4 \times \sin^{- 1} \left( \frac{x - 2}{2} \right) \right]_0^1 + \left[ \frac{1}{2}x\sqrt{4 - x^2} + \frac{1}{2} \times 4 \times \sin^{- 1} \left( \frac{x}{2} \right) \right]_1^2 \]
\[ = \left[ \frac{- \sqrt{3}}{2} + 2 \sin^{- 1} \left( \frac{- 1}{2} \right) \right] - \left[ 0 + 2 \sin^{- 1} \left( - 1 \right) \right] + \left( 0 - \frac{1}{2}\sqrt{3} \right) + 2\left\{ si n^{- 1} \left( 1 \right) - si n^{- 1} \left( \frac{1}{2} \right) \right\}\]
\[ = \frac{- \sqrt{3}}{2} + 2 \sin^{- 1} \left( \frac{- 1}{2} \right) - 0 - 2 \sin^{- 1} \left( - 1 \right) + 0 - \frac{1}{2}\sqrt{3} + 2si n^{- 1} \left( 1 \right) - 2si n^{- 1} \left( \frac{1}{2} \right)\]
\[ = - \sqrt{3} - 4 \sin^{- 1} \left( \frac{1}{2} \right) + 4 \sin^{- 1} \left( 1 \right)\]
\[ = - \sqrt{3} - 4 \times \frac{\pi}{6} + 4 \times \frac{\pi}{2}\]
\[ = - \sqrt{3} - \frac{2\pi}{3} + 2\pi\]
\[ = \frac{4\pi}{3} - \sqrt{3}\]
\[\text{ Now, From equation }\left( 1 \right)\]
\[\text{ Shaded area }= 2 \times\text{ area }\left(\text{ OBAO }\right) = 2\left( \frac{4\pi}{3} - \sqrt{3} \right) = \left( \frac{8\pi}{3} - 2\sqrt{3} \right)\text{ sq units }\]