Question

Find the area bounded by the curves \[ x^2 + y^2 = 4 \quad \text{and} \quad x^2 + (y-2)^2 = 4. \]

• A. \( \dfrac{8\pi}{3} - 2\sqrt{3} \) Sq. units
• B. \( \dfrac{8\pi}{3} + \sqrt{3} \) Sq. units
• C. \( \dfrac{4\pi}{3} - 2\sqrt{3} \) Sq. units
• D. \( \dfrac{4\pi}{3} + 2\sqrt{3} \) Sq. units

Hint:

For overlapping circles of equal radii, memorizing the standard common-area formula saves time.

Answer 1

The Correct Option is A

Step 1: Identify the circles.
The equations represent two circles of radius \(2\). First circle is centered at \( (0,0) \), second at \( (0,2) \).
Step 2: Find the distance between centers.
\[ d = \sqrt{(0-0)^2 + (2-0)^2} = 2 \]
Step 3: Use the formula for common area of two equal circles.
For two circles of radius \( r \) with distance \( d \) between centers, \[ \text{Common area} = 2r^2 \cos^{-1}\!\left(\frac{d}{2r}\right) - \frac{d}{2}\sqrt{4r^2 - d^2} \]
Step 4: Substitute values.
Here \( r = 2 \), \( d = 2 \): \[ \text{Area} = 2(4)\cos^{-1}\!\left(\frac{1}{2}\right) - \frac{2}{2}\sqrt{16 - 4} \] \[ = 8\left(\frac{\pi}{3}\right) - \sqrt{12} = \frac{8\pi}{3} - 2\sqrt{3} \]
Step 5: Final Answer.
\[ \boxed{\dfrac{8\pi}{3} - 2\sqrt{3}} \text{ Sq. units} \]