In the XY–plane, the area, in sq. units, of the region defined by the inequalities... y≥x+4 and −4≤x²+y²+4(x−y)≤0 is

The correct option is 2π.

To find the area of the defined region, we first analyze the given inequalities step-by-step.

Step 1: Simplify the quadratic inequality
The second inequality is given by:
$$-4\le x^2+y^2+4(x-y)\le 0$$
We can rewrite the middle term, $x^2+y^2+4x-4y$, by completing the square for both the $x$ and $y$ terms:
$$x^2+4x={(x+2)}^2-4$$
$$y^2-4y={(y-2)}^2-4$$
Substituting these completed squares back into the inequality gives:
$$-4\le {(x+2)}^2-4+{(y-2)}^2-4\le 0$$
Simplifying the constants:
$$-4\le {(x+2)}^2+{(y-2)}^2-8\le 0$$
Adding $8$ to all parts of the inequality:
$$4\le {(x+2)}^2+{(y-2)}^2\le 8$$

This inequality describes the region between two concentric circles centered at $(-2,2)$, known as an annulus:
- The inner circle has a radius of $r_1=\sqrt{4}=2$.
- The outer circle has a radius of $r_2=\sqrt{8}=2\sqrt{2}$.

Step 2: Incorporate the linear boundary
The first inequality is:
$$y\ge x+4$$
The boundary line of this region is $y=x+4$. Let us check if this line passes through the common center of the circles, $(-2,2)$:
Substitute $x=-2$ into the line equation:
$$y=-2+4=2$$
Since $y=2$, the boundary line passes directly through the center $(-2,2)$ of the annulus.

Step 3: Calculate the area
Because the boundary line passes through the center of the concentric circles, it cuts the circular region (and the annulus) exactly in half. The inequality $y\ge x+4$ defines the half-plane above this line, which contains exactly half of the total area of the annulus.
First, we find the total area of the annulus:
$${\text{Area}}_{\text{annulus}}=\pi (r_2^2-r_1^2)$$
$${\text{Area}}_{\text{annulus}}=\pi (8-4)=4\pi$$
The area of the region defined by both inequalities is half of this total area:
$$\text{Required Area}=\frac{1}{2}\times 4\pi =2\pi$$