Question

Find the area of the minor segment of the circle $x^2+y^2=a^2$ cut off by the line $x=\frac{a}{2}$.

Answer 1

The equation of the circle is $x^2+y^2=a^2$.

Centre of the circle = (0, 0) and radius = a.

The line $x=\frac{a}{2}$ is parallel to y-axis and intersects the x-axis at $\left(\frac{a}{2},0\right)$.


Required area = Area of the shaded region

= 2 × Area of the region ABDA

$$\begin{gathered} =2\times {\int }_{\frac{a}{2}}^a{y}_{\mathrm{circle}}dx \\ =2{\int }_{\frac{a}{2}}^a\sqrt{a^2-x^2}dx \\ =2{\overline{)\left(\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}\right)}}_{\frac{a}{2}}^a \\ =2\left[\left(0+\frac{a^2}{2}{\sin }^{-1}1\right)-\left(\frac{a}{4}\times \sqrt{a^2-\frac{a^2}{4}}+\frac{a^2}{2}{\sin }^{-1}\frac{1}{2}\right)\right] \end{gathered}$$
$$\begin{gathered} =2\left(\frac{a^2}{2}\times \frac{\mathrm{\pi }}{2}-\frac{a}{4}\times \frac{\sqrt{3}a}{2}-\frac{a^2}{2}\times \frac{\mathrm{\pi }}{6}\right) \\ =\frac{a^2\mathrm{\pi }}{2}-\frac{\sqrt{3}{a}^2}{4}-\frac{a^2\mathrm{\pi }}{6} \\ =\frac{6a^2\mathrm{\pi }-3\sqrt{3}{a}^2-2a^2\mathrm{\pi }}{12} \\ =\frac{a^2}{12}\left(4\mathrm{\pi }-3\sqrt{3}\right)\mathrm{square}\mathrm{units} \end{gathered}$$