Let $R\left\{(x, y): x^2+y^2=1, x, y \in R\right\}$ be a relation in R. The relation R is :

symmetric

We have $R=\left\{(x, y): x^2+y^2=1 ; x, y \in R\right\}$

$4 \in abd(4) 2+(4) 2=32 \neq 1$       ∴  $(4,4) \notin R$

⇒  R is not reflexive.

Let $(x, y) \in R$       ∴  $x^2+y^2=1$

$\Rightarrow y^2+x^2=1 \Rightarrow(y, x) \in R$

⇒ R is symmetric

$(0,1),(1,0) \in R$ because

(0)2 + (1)2 = 1 and (1)2 + (0)2 = 1

Also $(0) 2+(0) 2=0 \neq 1$         ⇒  $(0,0) \notin R$.

⇒ R is not transitive.

Hence (2) is the correct answer.