Question:medium

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

Updated On: Mar 23, 2026
  • \(2\pi\)
  • \(3\pi\)
  • \(\pi\)
  • \(4\pi\)
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The Correct Option is A

Solution and Explanation

Consider the second inequality:

$-4 \le x^2 + y^2 + 4(x - y) \le 0$.

Rearranging the second inequality, we get:

$x^2 + y^2 + 4x - 4y \le 4$,
$x^2 + y^2 + 4x - 4y + 4 \le 8$,
$(x + 2)^2 + (y - 2)^2 \le 8$.

This describes a circle centered at $(-2, 2)$ with a radius of $\sqrt{8}$, which simplifies to $2\sqrt{2}$.
Now, let's incorporate the first inequality, $y \ge x + 4$. This represents the region above the line $y = x + 4$.
The area we are looking for is the portion of the circle that lies above this line.
Since the line $y = x + 4$ passes through the center of the circle, it divides the circle into two equal halves.
The total area of the circle is $\pi \times (2\sqrt{2})^2 = 8\pi$. Therefore, the area of the region is:

$\frac{8\pi}{2} = 4\pi$.

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

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