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

1.	Let \[P=\{(x,\,y)\|{{x}^{2}}+{{y}^{2}}=1,\,x,\,y\in R\}\]. Then P is
A.	Reflexive
B.	Symmetric
C.	Transitive
D.	Anti-symmetric
Answer» C. Transitive

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

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