Let Z be the set of all integers,A = {(x, y) | vedclass

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(B) First,find the set $A \cap B$:$A = \{(x, y) \in Z 	imes Z : (x-2)^{2} + y^{2} \leq 4\}$$B = \{(x, y) \in Z 	imes Z : x^{2} + y^{2} \leq 4\}$The

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$25$

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(B) First,find the set $A \cap B$:$A = \{(x, y) \in Z 	imes Z : (x-2)^{2} + y^{2} \leq 4\}$$B = \{(x, y) \in Z 	imes Z : x^{2} + y^{2} \leq 4\}$The integer points $(x, y)$ satisfying both inequalities are $(1, 0), (1, 1), (1, -1), (2, 0), (0, 0)$.Thus,$n(A \cap B) = 5$.Next,find the set $A \cap C$:$A = \{(x, y) \in Z 	imes Z : (x-2)^{2} + y^{2} \leq 4\}$$C = \{(x, y) \in Z 	imes Z : (x-2)^{2} + (y-2)^{2} \leq 4\}$The integer points $(x, y)$ satisfying both inequalities are $(2, 0), (2, 1), (2, 2), (1, 1), (3, 1)$.Thus,$n(A \cap C) = 5$.The total number of relations from $A \cap B$ to $A \cap C$ is given by $2^{n(A \cap B) 	imes n(A \cap C)} = 2^{5 	imes 5} = 2^{25}$.Comparing this with $2^{p}$,we get $p = 25$.

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