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Let $Z$ be the set of all integers,
$\mathrm{A}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{Z} \times \mathbb{Z}:(\mathrm{x}-2)^{2}+\mathrm{y}^{2} \leq 4\right\}$
$\mathrm{B}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{Z} \times \mathbb{Z}: \mathrm{x}^{2}+\mathrm{y}^{2} \leq 4\right\} \text { and }$
$\mathrm{C}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{Z} \times \mathbb{Z}:(\mathrm{x}-2)^{2}+(\mathrm{y}-2)^{2} \leq 4\right\}$
If the total number of relation from $\mathrm{A} \cap \mathrm{B}$ to $\mathrm{A} \cap \mathrm{C}$ is $2^{\mathrm{p}}$, then the value of $\mathrm{p}$ is :
$16$
$25$
$49$
$9$
Solution
$(x-2)^{2}+y^{2} \leq 4$
$x^{2}+y^{2} \leq 4$
No. of points common in $\mathrm{C}_{1} \,\&\, \mathrm{C}_{2}$ is $5 .$
$(0,0),(1,0),(2,0),(1,1),(1,-1)$
Similarly in $\mathrm{C}_{2} \& \mathrm{C}_{3}$ is $5$
No. of relations $=2^{5 \times 5}=2^{25}$.
