Match List-I with List-II
| List-I (Region) | List-II (Area in sq. units) | ||
| (A) | Region enclosed by the circle \(x^{2}+y^{2}=9\) | (I) | \(\displaystyle\frac{9\pi}{2}\) |
| (B) | \(\{(x, y): x^{2}+y^{2} \le 16,\ y \le 0\}\) | (II) | \(9\pi\) |
| (C) | \(\{(x, y): x^{2}+y^{2} \le 4,\ x \ge 0,\ y \ge 0\}\) | (III) | \(\pi\) |
| (D) | \(\{(x, y): x^{2}+y^{2} \le 36,\ 0 \le y \le x\}\) | (IV) | \(8\pi\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
Each region is a fraction of a disc \(x^{2}+y^{2} \le r^{2}\), whose full area is \(\pi r^{2}\).
(A) Region enclosed by the circle \(x^{2}+y^{2}=9\)
This matches (II) \(9\pi\).
\(r = 3,\ \ \pi r^{2} = 9\pi\)
(B) \(\{(x, y): x^{2}+y^{2} \le 16,\ y \le 0\}\)
This matches (IV) \(8\pi\).
It is the lower half of the disc of radius 4:
\( \displaystyle \frac{1}{2}\pi(4)^{2} = 8\pi \)
(C) \(\{(x, y): x^{2}+y^{2} \le 4,\ x \ge 0,\ y \ge 0\}\)
This matches (III) \(\pi\).
It is the first-quadrant quarter of the disc of radius 2:
\( \displaystyle \frac{1}{4}\pi(2)^{2} = \pi \)
(D) \(\{(x, y): x^{2}+y^{2} \le 36,\ 0 \le y \le x\}\)
This matches (I) \(\displaystyle\frac{9\pi}{2}\).
The condition \(0 \le y \le x\) keeps the points between the x-axis and the line y = x, which makes an angle of \(45^{\circ}\) with the x-axis. So the region is a sector of angle \(45^{\circ}\), one-eighth of the disc of radius 6:
\( \displaystyle \frac{1}{8}\pi(6)^{2} = \frac{9\pi}{2} \)