If R=(x, y): x, y ∈ Z, x^2+y^2 ≤ 4 is a relation in Z, then domain of…

We have $R=\left\{(x, y): x, y \in Z, x^2+y^2 \leq 4\right\}$

Let x = 0   ∴ $x^2+y^2 \leq 4 \Rightarrow y^2 \leq 4 \Rightarrow y=0, \pm 1, \pm 2$

Let $x= \pm 2 ∴ x^2+y^2 \leq 4 \Rightarrow y^2 \leq 0 \Rightarrow y=0$

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

$x=1$ and $x=-1$

as we are talking of all points

such that $x^2+y^2≤4$ (Not $x^2+y^2=4$)

therefore, $1,-1$ ∈ Domain of inequality

∴ Domain of R = $\{x:(x, y) \in R\}=\{0,-1,1,-2,2\}$

Hence (3) is the correct answer.