Question

Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side 2 cm. What is the area of the portion that is common to both the circles?

• A. $2(\pi– 2) cm^2 $
• B. $2(\pi – 1) cm^2 $
• C. $2\pi cm^2 $
• D. $(\pi – 2) cm^2 $
• E. $(\pi – 1) cm^2 $

Answer 1

The Correct Option is A

To solve this problem, we need to find out the area of the region common to both identical intersecting circles, given that their centers and intersection points form a square of side 2 cm. 

1. Since the centers of the circles and the intersection points form a square, the distance between the centers of the circles is 2 cm (the side of the square).
2. Because the circles are identical, the radius of each circle can be determined by recognizing that the distance between their centers, which forms the diagonal of the square, is equal to the radius \(r\) of the circles.
3. Applying the Pythagorean theorem to the square, we get the diagonal: \(r\sqrt{2} = \text{diagonal of the square} = 2\sqrt{2}\), thus, \(r = 2\).
4. To find the area of the common region, we calculate the area of the lune (region formed by two intersecting circles):
5. The formula for the area of the intersection of two circles (lune) in this symmetric case is: \(A = 2 \times (r^2 \cos^{-1}(\frac{d}{2r}) - \frac{d}{2}\sqrt{r^2 - (\frac{d}{2})^2})\), where \(d\) is the distance between centers, here \(d = 2\).
6. Substituting the values, \(r = 2\) and \(d = 2\), the expression becomes: \(A = 2 \times (2^2 \cos^{-1}(\frac{2}{4}) - 1 \times \sqrt{4 - 1})\).
7. Simplify this: \(\cos^{-1}(0.5) = \frac{\pi}{3}\) (since \(\cos(\frac{\pi}{3}) = 0.5\)),
8. The area is then: \(A = 2 \times (4 \times \frac{\pi}{3} - \sqrt{3})\),
9. Therefore, the area of the common portion is: \(= 2(\frac{4\pi}{3} - \sqrt{3})\).
10. Multiplying terms gives the answer as \(2(\pi - 2)\) cm², which matches with the given correct option.

Thus, the area of the region common to both circles is \(2(\pi - 2)\) cm².