Question

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

• A. \(2\pi\)
• B. \(3\pi\)
• C. \(\pi\)
• D. \(4\pi\)

Answer 1

The Correct Option is A

Consider the second inequality:

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

We can rewrite the second inequality:

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

This represents a circle centered at $(-2, 2)$ with radius $\sqrt{8} = 2\sqrt{2}$.
Now, combine the first inequality $y \ge x + 4$, which represents the region above the line $y = x + 4$.
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 $y = x + 4$ divides the circle into two equal parts.
The 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 $2\pi$.