Let P = { (x, y) | x^2 + y^2 = 1, x, y in R } | vedclass

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(B) relation $P$ on a set $R$ is defined by the condition $x^2 + y^2 = 1$.$1$. Reflexive: For $P$ to be reflexive,$(x, x) \in P$ must hold for all $x

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Symmetric

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(B) relation $P$ on a set $R$ is defined by the condition $x^2 + y^2 = 1$.$1$. Reflexive: For $P$ to be reflexive,$(x, x) \in P$ must hold for all $x \in R$. This implies $x^2 + x^2 = 1$,or $2x^2 = 1$,which is not true for all $x \in R$ (e.g.,$x = 0$). Thus,$P$ is not reflexive.$2$. Symmetric: For $P$ to be symmetric,if $(x, y) \in P$,then $(y, x) \in P$ must hold. Given $x^2 + y^2 = 1$,it follows that $y^2 + x^2 = 1$. Therefore,$(y, x) \in P$. Thus,$P$ is symmetric.$3$. Transitive: For $P$ to be transitive,if $(x, y) \in P$ and $(y, z) \in P$,then $(x, z) \in P$ must hold. Let $x = 1, y = 0, z = 1$. Then $(1, 0) \in P$ $(1^2 + 0^2 = 1)$ and $(0, 1) \in P$ $(0^2 + 1^2 = 1)$. However,$(1, 1) 
otin P$ because $1^2 + 1^2 = 2 
eq 1$. Thus,$P$ is not transitive.Conclusion: The relation $P$ is symmetric.

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

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