CAT 2024 Slot 1 QA Question 10

First, the question mentions two inequalities: y ≥ x + 4 and -4 ≤ x² + y² + 4(x - y) ≤ 0. We find the area of the region that satisfies both these conditions. The options are multiples of π, so I suspect the region might be a circle or something related.

Let me start with the second inequality because it looks more complicated: -4 ≤ x² + y² + 4(x - y) ≤ 0. Hmm, maybe I can rewrite this equation to recognize it as a circle. the general equation of a circle is x² + y² + Dx + Ey + F = 0, which can be rewritten in standard form by completing the squares.

So, Taking the expression x² + y² + 4x - 4y. I can group the x terms and y terms together:

x² + 4x + y² - 4y.

Now, I'll complete the square for both x and y.

For the x terms: x² + 4x. To complete the square, I take half of 4, which is 2, square it to get 4, and add it. Similarly, for the y terms: y² - 4y. Half of -4 is -2, squaring it gives 4, so I add 4 there too.

So, adding these squares:

(x² + 4x + 4) + (y² - 4y + 4) = (x + 2)² + (y - 2)².

But wait, I added 4 and 4, which is 8, to both sides. So the original inequality becomes:

-4 ≤ (x + 2)² + (y - 2)² - 8 ≤ 0.

The original expression is x² + y² + 4x - 4y, which is equal to (x + 2)² + (y - 2)² - 8. So, substituting back into the inequality:

-4 ≤ (x + 2)² + (y - 2)² - 8 ≤ 0.

Let me add 8 to all parts of the inequality to isolate the squared terms:

-4 + 8 ≤ (x + 2)² + (y - 2)² ≤ 0 + 8.

So that simplifies to:

4 ≤ (x + 2)² + (y - 2)² ≤ 8.

Wait, but that can't be right because the left side is 4 and the right side is 8. Hmm, maybe I made a mistake in the inequality direction. Let me double-check.

Wait, the original inequality is -4 ≤ x² + y² + 4x - 4y ≤ 0. So when I complete the square, I have:

x² + y² + 4x - 4y = (x + 2)^2 - 4 + (y - 2)^2 - 4 = (x + 2)^2 + (y - 2)^2 - 8.

So substituting back, the inequality becomes:

-4 ≤ (x + 2)^2 + (y - 2)^2 - 8 ≤ 0.

Now, adding 8 to all parts:

-4 + 8 ≤ (x + 2)^2 + (y - 2)^2 ≤ 0 + 8.

Which is 4 ≤ (x + 2)^2 + (y - 2)^2 ≤ 8.

So the region defined by this inequality is the set of points (x, y) such that the distance squared from (-2, 2) is between 4 and 8. So, it's an annular region between two circles centered at (-2, 2) with radii 2 and 2√2, since 4 is 2² and 8 is (2√2)².

So, the area between these two circles would be the area of the larger circle minus the area of the smaller one. The area of a circle is πr², so the area between them is π( (2√2)^2 ) - π(2^2 ) = π(8) - π(4) = 4π. But wait, that's the area between the two circles, but I also have the other inequality y ≥ x + 4 to consider.

So, the region we're interested in is the intersection of the annular region between the two circles and the region where y is greater than or equal to x + 4.

The center of the circles is at (-2, 2). The line y = x + 4 is a straight line with a slope of 1 and y-intercept at 4. where this line intersects the circles.

First, let's find the points where y = x + 4 intersects the inner circle (radius 2) and the outer circle (radius 2√2).

Starting with the inner circle: (x + 2)^2 + (y - 2)^2 = 4.

Substituting y = x + 4 into this equation:

(x + 2)^2 + ( (x + 4) - 2 )^2 = 4.

Simplify:

(x + 2)^2 + (x + 2)^2 = 4.

Which is 2*(x + 2)^2 = 4.

Divide both sides by 2: (x + 2)^2 = 2.

Taking square roots: x + 2 = ±√2.

So, x = -2 ± √2.

Then, y = x + 4, so when x = -2 + √2, y = (-2 + √2) + 4 = 2 + √2.

Similarly, when x = -2 - √2, y = (-2 - √2) + 4 = 2 - √2.

So the points of intersection with the inner circle are (-2 + √2, 2 + √2) and (-2 - √2, 2 - √2).

Now, let's check if these points lie on the line y = x + 4. Well, yes, by construction.

Now, let's check the outer circle: (x + 2)^2 + (y - 2)^2 = 8.

Substituting y = x + 4:

(x + 2)^2 + (x + 4 - 2)^2 = 8.

Simplify:

(x + 2)^2 + (x + 2)^2 = 8.

Which is 2*(x + 2)^2 = 8.

Divide by 2: (x + 2)^2 =