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

We begin by recalling that a relation from one set to another is simply a subset of the Cartesian product of the two sets. If the first set has $$m$$ elements and the second has $$n$$ elements, then the product has $$m\,n$$ ordered pairs, and for every ordered pair we have a choice of either “include it in the relation’’ or “do not include it.’’ Hence the total number of possible relations is $$2^{m\,n}.$$ In our problem we therefore need the sizes of the two intersections $$A\cap B$$ and $$A\cap C.$$ After that we will substitute them into the formula Number of relations $$=2^{|A\cap B|\,|A\cap C|}.$$

The three sets are defined inside the lattice $$\mathbb Z\times\mathbb Z$$ by the following inequalities:

$$\begin{aligned} A &:=\{(x,y)\in\mathbb Z\times\mathbb Z:(x-2)^2+y^2\le 4\},\\[2mm] B &:=\{(x,y)\in\mathbb Z\times\mathbb Z:x^2+y^2\le 4\},\\[2mm] C &:=\{(x,y)\in\mathbb Z\times\mathbb Z:(x-2)^2+(y-2)^2\le 4\}. \end{aligned}$$

Set $$A$$ is the collection of all integer points inside or on the circle of radius $$2$$ centred at $$(2,0).$$ Likewise, $$B$$ is the radius-$$2$$ circle centred at the origin, and $$C$$ is the radius-$$2$$ circle centred at $$(2,2).$$

Counting the points in $$A\cap B$$. A point $$(x,y)$$ must satisfy both

$$\begin{cases} (x-2)^2+y^2\le4,\\ x^2+y^2\le4. \end{cases}$$

Because $$x^2+y^2\le4,$$ the coordinate $$x$$ can only be $$-2,-1,0,1,2.$$ On the other hand, from $$(x-2)^2+y^2\le4$$ we have $$0\le x\le4.$$ Taking the intersection of these possibilities gives $$x\in\{0,1,2\}.$$ We examine each value separately.

For $$x=0$$:

$$x^2+y^2\le4\;\Rightarrow\;y^2\le4\;\Rightarrow\;y=-2,-1,0,1,2.$$ $$(x-2)^2+y^2=(0-2)^2+y^2=4+y^2\le4\;\Rightarrow\;y^2\le0\;\Rightarrow\;y=0.$$ So only the point $$(0,0)$$ remains.

For $$x=1$$:

$$x^2+y^2=1+y^2\le4\;\Rightarrow\;y^2\le3\;\Rightarrow\;y=-1,0,1.$$ $$(x-2)^2+y^2=(1-2)^2+y^2=1+y^2\le4\;\Rightarrow\;y=-1,0,1.$$ All three survive, giving the points $$(1,-1),(1,0),(1,1).$$

For $$x=2$$:

$$x^2+y^2=4+y^2\le4\;\Rightarrow\;y^2\le0\;\Rightarrow\;y=0.$$ $$(x-2)^2+y^2=0+y^2\le4\;\Rightarrow\;y=-2,-1,0,1,2.$$ The common value is $$y=0,$$ so we get the single point $$(2,0).$$

Collecting everything,

$$|A\cap B|=1+3+1=5.$$

Counting the points in $$A\cap C$$. A point lies in both $$A$$ and $$C$$ precisely when

$$\begin{cases} (x-2)^2+y^2\le4,\\ (x-2)^2+(y-2)^2\le4. \end{cases}$$

It is convenient to set $$u=x-2.$$ Then $$u$$ is an integer with $$u^2\le4,$$ so $$u\in\{-2,-1,0,1,2\}.$$ In these new variables the inequalities become

$$\begin{cases} u^2+y^2\le4,\\ u^2+(y-2)^2\le4. \end{cases}$$

For every fixed $$u$$ write $$R=4-u^2.$$ Both squares must not exceed $$R,$$ so we need simultaneously

$$|y|\le\sqrt R\quad\text{and}\quad|y-2|\le\sqrt R.$$

For $$u=\pm2$$: $$u^2=4\Rightarrow R=0.$$ Then $$y=0$$ from the first absolute-value condition and $$y=2$$ from the second, an impossibility. Hence no points occur for $$u=\pm2.$$

For $$u=\pm1$$: $$u^2=1\Rightarrow R=3$$, $$\;\sqrt R\approx1.732.$$ First condition gives $$y=-1,0,1;$$ second gives $$y-2=-1,0,1\;\Rightarrow\;y=1,2,3.$$ Their intersection is the single value $$y=1.$$ Thus we obtain the two points $$u=-1\Rightarrow x=1:\;(1,1)$$ and $$u=1\Rightarrow x=3:\;(3,1).$$

For $$u=0$$: $$u^2=0\Rightarrow R=4,\;\sqrt R=2.$$ First condition yields $$y=-2,-1,0,1,2;$$ second yields $$y=0,1,2,3,4.$$ The overlap is $$y=0,1,2.$$ These correspond to the three points $$(2,0),(2,1),(2,2).$$

Hence

$$|A\cap C|=2+3=5.$$

Counting the relations. We have found

$$|A\cap B|=5,\qquad |A\cap C|=5.$$

The Cartesian product $$(A\cap B)\times(A\cap C)$$ therefore contains $$5\times5=25$$ ordered pairs. According to the formula stated at the start, the number of relations from $$A\cap B$$ to $$A\cap C$$ is

$$2^{25}.$$

Thus $$p=25.$$ Among the given options, this matches Option A.

Hence, the correct answer is Option A.