Key Concepts and Formulas

• Integer solutions to inequalities: Finding integer pairs $(x, y)$ that satisfy inequalities defining regions in the Cartesian plane.
• Intersection of sets: $A \cap B$ represents the set of elements common to both $A$ and $B$.
• Number of relations: The number of relations from a set $X$ to a set $Y$ is $2^{|X| \cdot |Y|}$, where $|X|$ and $|Y|$ denote the number of elements in sets $X$ and $Y$, respectively.

Step-by-Step Solution

Step 1: Determine the elements of set A

Set A is defined by $(x-2)^2 + y^2 \le 4$. This represents all integer points $(x,y)$ inside or on the circle centered at $(2,0)$ with radius 2. We need to find all integer pairs $(x,y)$ that satisfy this inequality.

When $y=0$, we have $(x-2)^2 \le 4$, which means $-2 \le x-2 \le 2$, so $0 \le x \le 4$. This gives us the points $(0,0), (1,0), (2,0), (3,0), (4,0)$.

When $y=1$, we have $(x-2)^2 + 1 \le 4$, so $(x-2)^2 \le 3$, which means $-\sqrt{3} \le x-2 \le \sqrt{3}$. Since $x$ is an integer, $-1 \le x-2 \le 1$, so $1 \le x \le 3$. This gives us the points $(1,1), (2,1), (3,1)$. When $y=-1$, we have $(x-2)^2 + 1 \le 4$, so $(x-2)^2 \le 3$, which means $-\sqrt{3} \le x-2 \le \sqrt{3}$. Since $x$ is an integer, $-1 \le x-2 \le 1$, so $1 \le x \le 3$. This gives us the points $(1,-1), (2,-1), (3,-1)$.

When $y=2$, we have $(x-2)^2 + 4 \le 4$, so $(x-2)^2 \le 0$, which means $x-2=0$, so $x=2$. This gives us the point $(2,2)$. When $y=-2$, we have $(x-2)^2 + 4 \le 4$, so $(x-2)^2 \le 0$, which means $x-2=0$, so $x=2$. This gives us the point $(2,-2)$.

Therefore, $A = \{ (0,0), (1,0), (2,0), (3,0), (4,0), (1,1), (2,1), (3,1), (1,-1), (2,-1), (3,-1), (2,2), (2,-2) \}$.

Step 2: Determine the elements of set B

Set B is defined by $x^2 + y^2 \le 4$. This represents all integer points $(x,y)$ inside or on the circle centered at $(0,0)$ with radius 2. We need to find all integer pairs $(x,y)$ that satisfy this inequality.

When $y=0$, we have $x^2 \le 4$, which means $-2 \le x \le 2$. This gives us the points $(-2,0), (-1,0), (0,0), (1,0), (2,0)$.

When $y=1$, we have $x^2 + 1 \le 4$, so $x^2 \le 3$, which means $-\sqrt{3} \le x \le \sqrt{3}$. Since $x$ is an integer, $-1 \le x \le 1$. This gives us the points $(-1,1), (0,1), (1,1)$. When $y=-1$, we have $x^2 + 1 \le 4$, so $x^2 \le 3$, which means $-\sqrt{3} \le x \le \sqrt{3}$. Since $x$ is an integer, $-1 \le x \le 1$. This gives us the points $(-1,-1), (0,-1), (1,-1)$.

When $y=2$, we have $x^2 + 4 \le 4$, so $x^2 \le 0$, which means $x=0$. This gives us the point $(0,2)$. When $y=-2$, we have $x^2 + 4 \le 4$, so $x^2 \le 0$, which means $x=0$. This gives us the point $(0,-2)$.

Therefore, $B = \{ (-2,0), (-1,0), (0,0), (1,0), (2,0), (-1,1), (0,1), (1,1), (-1,-1), (0,-1), (1,-1), (0,2), (0,-2) \}$.

Step 3: Determine the elements of set C

Set C is defined by $(x-2)^2 + (y-2)^2 \le 4$. This represents all integer points $(x,y)$ inside or on the circle centered at $(2,2)$ with radius 2. We need to find all integer pairs $(x,y)$ that satisfy this inequality. This is simply a translation of the set B by the vector $\begin{pmatrix} 2 \\ 2 \end{pmatrix}$.

Applying the translation to the elements of B, we get: $C = \{ (0+2,0+2), (-1+2,0+2), (0+2,0+2), (1+2,0+2), (2+2,0+2), (-1+2,1+2), (0+2,1+2), (1+2,1+2), (-1+2,-1+2), (0+2,-1+2), (1+2,-1+2), (0+2,2+2), (0+2,-2+2) \} = \{ (2,2), (1,2), (2,2), (3,2), (4,2), (1,3), (2,3), (3,3), (1,1), (2,1), (3,1), (2,4), (2,0) \}$. Removing duplicates: $C = \{ (2,2), (1,2), (3,2), (4,2), (1,3), (2,3), (3,3), (1,1), (2,1), (3,1), (2,4), (2,0) \}$.

Step 4: Determine the intersection of A and B

$A \cap B = \{ (0,0), (1,0), (2,0), (3,0), (4,0), (1,1), (2,1), (3,1), (1,-1), (2,-1), (3,-1), (2,2), (2,-2) \} \cap \{ (-2,0), (-1,0), (0,0), (1,0), (2,0), (-1,1), (0,1), (1,1), (-1,-1), (0,-1), (1,-1), (0,2), (0,-2) \} = \{ (0,0), (1,0), (2,0), (1,1) \}$. Thus, $|A \cap B| = 4$.

Step 5: Determine the intersection of A and C

$A \cap C = \{ (0,0), (1,0), (2,0), (3,0), (4,0), (1,1), (2,1), (3,1), (1,-1), (2,-1), (3,-1), (2,2), (2,-2) \} \cap \{ (2,2), (1,2), (3,2), (4,2), (1,3), (2,3), (3,3), (1,1), (2,1), (3,1), (2,4), (2,0) \} = \{ (1,1), (2,1), (3,1), (2,0) \}$. Thus, $|A \cap C| = 4$.

Step 6: Determine the number of relations from A $\cap$ B to A $\cap$ C

The number of relations from $A \cap B$ to $A \cap C$ is $2^{|A \cap B| \cdot |A \cap C|} = 2^{4 \cdot 4} = 2^{16}$. We are given that this is $2^p$, so $p = 16$.

Common Mistakes & Tips

• Carefully list out all integer points satisfying the inequalities. It's easy to miss a point or include a non-integer point.
• Remember that the number of relations from set X to set Y is $2^{|X| \cdot |Y|}$, not $2^{|X| + |Y|}$.
• When finding intersections, double-check each element to make sure it belongs to both sets.

Summary

The problem involves finding the intersection of sets of integer points within circular regions and then calculating the number of relations between these intersections. After carefully listing the elements of each set and their intersections, we found that $|A \cap B| = 4$ and $|A \cap C| = 4$. The number of relations from $A \cap B$ to $A \cap C$ is $2^{4 \cdot 4} = 2^{16}$, which implies that $p=16$.

Final Answer

The final answer is \boxed{16}, which corresponds to option (A).