1. Interpret the sets

We are given integer lattice points in three disks:

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

We need the number of relations from $A\cap B$ to $A\cap C$.

If $|A\cap B|=m$ and $|A\cap C|=n$, then the number of relations from $A\cap B$ to $A\cap C$ is

$2^{mn}$

because any relation is any subset of $(A\cap B)\times (A\cap C)$.

The question says this number is $2^p$, so

$p=mn$

Thus we only need to find the sizes of $A\cap B$ and $A\cap C$.

2. Find $A\cap B$

A point $(x,y)$ belongs to $A\cap B$ if it satisfies both:

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

First list all integer points in $B$:

$x^2+y^2\le 4$

The integer points are:

• $y=0$: $x=-2,-1,0,1,2$
• $y=1$: $x=-1,0,1$
• $y=-1$: $x=-1,0,1$
• $y=2$: $x=0$
• $y=-2$: $x=0$

So

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

Now check which of these also lie in $A$, i.e.

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

• $(-2,0)$: $( -4)^2=16>4$ → no
• $(-1,0)$: $(-3)^2=9>4$ → no
• $(0,0)$: $(-2)^2=4$ → yes
• $(1,0)$: $(-1)^2=1$ → yes
• $(2,0)$: $0$ → yes
• $(-1,1)$: $9+1=10$ → no
• $(0,1)$: $4+1=5$ → no
• $(1,1)$: $1+1=2$ → yes
• $(-1,-1)$: $9+1=10$ → no
• $(0,-1)$: $4+1=5$ → no
• $(1,-1)$: $1+1=2$ → yes
• $(0,2)$: $4+4=8$ → no
• $(0,-2)$: $4+4=8$ → no

Hence

$A\cap B=\{(0,0),(1,0),(2,0),(1,1),(1,-1)\}$

So,

$|A\cap B|=5$

3. Find $A\cap C$

A point $(x,y)$ belongs to $A\cap C$ if it satisfies both:

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

Let

$u=x-2$

Then the conditions become:

$u^2+y^2\le 4$ $u^2+(y-2)^2\le 4$

This is the same geometry as for $A\cap B$, just shifted by 2 units in the $x$-direction. So it will also have 5 points. Let us list them explicitly.

Integer points in $A$ are centered at $(2,0)$ with radius $2$:

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

Now test for membership in $C$:

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

• $(0,0)$: $4+4=8$ → no
• $(1,0)$: $1+4=5$ → no
• $(2,0)$: $0+4=4$ → yes
• $(3,0)$: $1+4=5$ → no
• $(4,0)$: $4+4=8$ → no
• $(1,1)$: $1+1=2$ → yes
• $(2,1)$: $0+1=1$ → yes
• $(3,1)$: $1+1=2$ → yes
• $(1,-1)$: $1+9=10$ → no
• $(2,-1)$: $0+9=9$ → no
• $(3,-1)$: $1+9=10$ → no
• $(2,2)$: $0+0=0$ → yes
• $(2,-2)$: $0+16=16$ → no

Hence

$A\cap C=\{(2,0),(1,1),(2,1),(3,1),(2,2)\}$

So,

$|A\cap C|=5$

4. Number of relations

If $X=A\cap B$ and $Y=A\cap C$, then

$|X|=5, \qquad |Y|=5$

Therefore,

$|X\times Y|=5\cdot 5=25$

Number of relations from $X$ to $Y$ is

$2^{|X\times Y|}=2^{25}$

Given this is $2^p$, we get

$p=25$

5. Check options

• A: $16$ ❌
• B: $25$ ✅
• C: $49$ ❌
• D: $9$ ❌

So the correct option is B.