Question

Smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2 is
(a) 2 (π − 2)
(b) π − 2
(c) 2π − 1
(d) 2 (π + 2)

Answer 1

(b) π − 2



We have, x² + y² = 4 represents a circle with centre at O(0,0) and radius 2
x + y = 2 represents a straight line cutting the x-axis at A(2, 0) and y axis at B(0, 2)

Thus , A (2,0) and B(0,2) are also the points of intersection of the straight line and the circle

Smaller area enclosed by the curve and straight line is the shaded area

$$\begin{gathered} \mathrm{Shaded}\mathrm{area}\left(ABCA\right) \\ =\mathrm{area}\left(OBCA\right)-\mathrm{area}\left(OBAO\right) \\ ={\int }_0^2\sqrt{4-x^2}dx-{\int }_0^2\left(2-x\right)dx\left[\because x^2+y^2=4\Rightarrow y=\sqrt{4-x^2} \\ \mathrm{and}x+y=2\Rightarrow y=2-x\right] \\ ={\int }_0^2\left[\left(\sqrt{4-x^2}\right)+x-2\right]dx \\ ={\left[\frac{1}{2}x\sqrt{4-x^2}+\frac{1}{2}\times 4\times {\sin }^{-1}\left(\frac{x}{2}\right)+\left(\frac{x^2}{2}-2x\right)\right]}_0^2 \\ =\frac{1}{2}\times 2\sqrt{4-2^2}+2\times {\sin }^{-1}\left(\frac{2}{2}\right)+\left(\frac{2^2}{2}-2\times 2\right)-0 \\ =0+2\times \frac{\mathrm{\pi }}{2}+\left(2-4\right) \\ =\left(\mathrm{\pi }-2\right)\mathrm{sq}\mathrm{units} \end{gathered}$$