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$Let P = \{(x, y) ∣ x^{2} + y^{2} = 1, x, y \in R\} . Then, P is$

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Reflexive

Symmetric

Transitive

Anti-symmetric

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detailed solution

Correct option is B

Obviously, the relation is not reflexive and transitive but it is symmetric, because x2+y2=1⇒y2+x2=1

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Let P=(x,y)∣x2+y2=1,x,y∈R . Then, P is

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$Let P = \{(x, y) ∣ x^{2} + y^{2} = 1, x, y \in R\} . Then, P is$