Question

The area of the smaller segment cut off from the circle ${\mathrm{x}}^2+{\mathrm{y}}^2=9$ by $\mathrm{x}=1$ is

• A. $\left[\left(\frac{1}{2}\right)\left(9{\sec }^{-1}\left(3\right)-\sqrt{8}\right)\right]\mathrm{sq}\mathrm{units}$
• B. $\left(9{\sec }^{-1}\left(3\right)-\sqrt{8}\right)\mathrm{sq}\mathrm{units}$
• C. $\left(\sqrt{8}-9{\sec }^{-1}\left(3\right)\right)\mathrm{sq}\mathrm{units}$
• D. None of these

Answer 1

The correct option is B

$\left(9{\sec }^{-1}\left(3\right)-\sqrt{8}\right)\mathrm{sq}\mathrm{units}$

Explanation for the correct option:

Finding the area of the smaller segment:

Given, equation of circle ${\mathrm{x}}^2+{\mathrm{y}}^2=9$

Area of the smaller part cuts by $\mathrm{x}=1$

Plotting both the curves,

The equation of the circle is in blue curve while the equation of straight line is in green curve.

So, the right-sided portion of the circle cut by the straight line is the required area.

Thus, the required area is

$$\begin{gathered} =2{\int }_1^3\sqrt{9-{\mathrm{x}}^2}d\mathrm{x}\mathbf{}\mathbf{\left(}\mathbf{\text{using integration by parts}}\mathbf{\right)} \\ =2\times \frac{1}{2}{\left[\mathrm{x}\sqrt{9-{\mathrm{x}}^2}+9{\sin }^{-1}\left(\frac{1}{3}\right)\right]}_1^3\left(\text{as we know},\int \sqrt{a^2-x^2}dx=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}si{n}^{-1}\left(\frac{x}{a}\right)\right) \\ =\left[9\times \frac{\mathrm{\pi }}{2}-\sqrt{8}-9{\sin }^{-1}\left(\frac{1}{3}\right)\right] \\ =\left[9\left(\frac{\mathrm{\pi }}{2}-{\sin }^{-1}\left(\frac{1}{3}\right)\right)-\sqrt{8}\right] \\ =\left[9\left({\cos }^{-1}\left(\frac{1}{3}\right)\right)-\sqrt{8}\right] \\ =\left[9{\sec }^{-1}\left(3\right)-\sqrt{8}\right]\mathrm{sq}.\mathrm{units} \end{gathered}$$

Hence, option (B) is the correct answer.