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Question

Quantitative Aptitude Question on Coordinate Geometry

In the XYXY-plane, the area, in sq. units, of the region defined by the inequalities yx+4y \ge x + 4 and 4x2+y2+4(xy)0-4 \le x^2 + y^2 + 4(x - y) \le 0 is

A

2π2\pi

B

3π3\pi

C

π\pi

D

4π4\pi

Answer

2π2\pi

Explanation

Solution

Consider the second inequality:

4x2+y2+4(xy)0-4 \le x^2 + y^2 + 4(x - y) \le 0.

We can rewrite the second inequality:

x2+y2+4x4y4x^2 + y^2 + 4x - 4y \le 4,
x2+y2+4x4y+48x^2 + y^2 + 4x - 4y + 4 \le 8,
(x+2)2+(y2)28(x + 2)^2 + (y - 2)^2 \le 8.

This represents a circle centered at (2,2)(-2, 2) with radius 8=22\sqrt{8} = 2\sqrt{2}.
Now, combine the first inequality yx+4y \ge x + 4, which represents the region above the line y=x+4y = x + 4.
The area of the region is the area of the circle segment cut off by the line.
This can be calculated as half of the circle, since the line y=x+4y = x + 4 divides the circle into two equal parts.
The area of the circle is π×(22)2=8π\pi \times (2\sqrt{2})^2 = 8\pi. Therefore, the area of the region is:

8π2=4π\frac{8\pi}{2} = 4\pi.

The area defined by the inequalities is 2π2\pi.