Let Z be the set of all integers, A = {(x, y) ∈ Z × Z : (x - 2)² + y² ≤ 4}, B = {(x, y) ∈ Z × Z : x²

Q: Wait, why are we only looking at integer points? The inequalities define regions, not just points.

That is a fantastic observation! You are absolutely right that the inequalities define continuous regions. However, the problem explicitly states that $A$, $B$, and $C$ are subsets of $\mathbb{Z} \times \mathbb{Z}$. This means we are restricted to the 'lattice points'—the points where both $x$ and $y$ are integers. We ignore the continuous space between the grid lines because the sets themselves are defined only over the integers.

Q: How can I be sure I haven't missed any points in the intersection?

The best way to be certain is to be systematic. For a circle like $x^2 + y^2 \le 4$, the possible integer values for $x$ are limited to $\{-2, -1, 0, 1, 2\}$. For each $x$, calculate the range of possible $y$ values using the inequality $y^2 \le 4 - x^2$. By testing each $x$ value one by one, you guarantee that you won't miss any valid integer pairs $(x, y)$.

Q: Why is the number of relations $2^{n(X) \times n(Y)}$? I thought it was just $n(X) \times n(Y)$.

Remember that a relation from set $X$ to set $Y$ is defined as a *subset* of the Cartesian product $X \times Y$. The total number of elements in the Cartesian product $X \times Y$ is $n(X) \times n(Y)$. Since a relation can be any subset of this product, and a set with $k$ elements has $2^k$ possible subsets, the total number of relations is $2^{n(X) \times n(Y)}$. It is a classic combinatorial result!

Question

Let $\mathbb{Z}$ be the set of all integers, $A = \{(x, y) \in \mathbb{Z} 	imes \mathbb{Z} : (x - 2)^2 + y^2 \le 4\}$, $B = \{(x, y) \in \mathbb{Z} 	imes \mathbb{Z} : x^2 + y^2 \le 4\}$ and $C = \{(x, y) \in \mathbb{Z} 	imes \mathbb{Z} : (x - 2)^2 + (y - 2)^2 \le 4\}$. If the total number of relation from $A \cap B$ to $A \cap C$ is $2^p$, then the value of $p$ is :

Options:

(A) 16

(B) 25

(C) 49

(D) 9

Sigma Percentile · Accepted Answer

### The Geometry of Integers: A Journey into Sets and Relations

Welcome, fellow explorer of the mathematical landscape. Today, we are not just solving a problem; we are embarking on a journey through the elegant world of coordinate geometry and combinatorics.

The JEE Advanced syllabus often presents problems that seem like a dense forest of inequalities, but once you clear the brush, you find a beautiful, structured path. Let us break down this problem together.

### Phase 1: Visualizing the Circles

We are given three sets, $A$, $B$, and $C$, defined by circular inequalities. Specifically, we have:

$$A = \{(x, y) \in \mathbb{Z} 	imes \mathbb{Z} : (x - 2)^2 + y^2 \le 4\}$$

$$B = \{(x, y) \in \mathbb{Z} 	imes \mathbb{Z} : x^2 + y^2 \le 4\}$$

$$C = \{(x, y) \in \mathbb{Z} 	imes \mathbb{Z} : (x - 2)^2 + (y - 2)^2 \le 4\}$$

Imagine a standard Cartesian grid. Set $B$ is a solid circle centered at the origin $(0, 0)$ with a radius of $2$. Set $A$ is an identical circle, but its center is shifted to $(2, 0)$.

Set $C$ is another identical circle, shifted to $(2, 2)$. The crucial constraint here is that we are working in $\mathbb{Z} 	imes \mathbb{Z}$. We are not interested in the entire shaded area of these circles, but only the specific points where the grid lines intersect—the integer coordinates.

### Phase 2: The Hunt for Points in $A \cap B$

Our first task is to find the intersection $A \cap B$. We need points $(x, y)$ that satisfy both $x^2 + y^2 \le 4$ and $(x - 2)^2 + y^2 \le 4$.

Let us test the integer values for $x$ systematically. If $x = 0$, the first inequality gives $y^2 \le 4$, and the second gives $(-2)^2 + y^2 \le 4$, which simplifies to $4 + y^2 \le 4$, implying $y = 0$. So, $(0, 0)$ is our first point.

If $x = 1$, the first inequality gives $1 + y^2 \le 4 \Rightarrow y^2 \le 3$, so $y \in \{-1, 0, 1\}$. The second inequality gives $(1 - 2)^2 + y^2 \le 4 \Rightarrow 1 + y^2 \le 4$, which is the same condition. Thus, we have three points: $(1, -1), (1, 0), (1, 1)$.

If $x = 2$, the first inequality gives $4 + y^2 \le 4 \Rightarrow y = 0$. The second gives $0 + y^2 \le 4$, which is satisfied by $y = 0$. So, $(2, 0)$ is our fifth point. We have found $n(A \cap B) = 5$.

### Phase 3: The Symmetry of $A \cap C$

Now, let us turn our attention to $A \cap C$. We need points satisfying $(x - 2)^2 + y^2 \le 4$ and $(x - 2)^2 + (y - 2)^2 \le 4$.

Notice the symmetry! If we let $X = x - 2$ and $Y = y - 1$, the equations become $X^2 + (Y+1)^2 \le 4$ and $X^2 + (Y-1)^2 \le 4$. This is just a shifted version of our previous intersection.

By testing the coordinates, we find the points are $(2, 0), (2, 1), (2, 2), (1, 1), (3, 1)$. Just like before, we have exactly five points. So, $n(A \cap C) = 5$. The geometry is consistent, and the symmetry holds.

### Phase 4: The Combinatorial Climax

We have reached the final stage. We need to find the number of relations from $X = A \cap B$ to $Y = A \cap C$. A relation is a subset of the Cartesian product $X 	imes Y$.

The number of elements in the Cartesian product is $n(X) 	imes n(Y) = 5 	imes 5 = 25$. The total number of subsets of a set with $25$ elements is $2^{25}$.

The problem tells us this value is $2^p$. Therefore, $2^p = 2^{25}$, which leads us directly to **$p = 25$**.

It is a beautiful result, isn't it? We started with circles and ended with a simple exponent. This is the essence of JEE Advanced—taking a complex geometric setup and distilling it into a fundamental principle.

Select Answer:

Visualized Solution

Understanding the Coordinate Space

Visualizing Sets $A$ and $B$

Integer Points in $A \cap B$

Visualizing Set $C$

Integer Points in $A \cap C$

Total Number of Relations

Calculating the Final Value

The Sigma Insight: Cartesian Product of Sets

The Geometry of Integers

Phase 1

Phase 2

Phase 3

Phase 4

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