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If $\rho=\left\{(x, y) \mid x^{2}+y^{2}=1 ; x, y \in \mathbb{R}\right\}$. Then, $\rho$ is
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Verified Answer
The correct answer is:
symmetric
Obviously, the relation is not reflexive and transitive, but it is symmetric, because $x^{2}+x^{2}=2 x^{2} \neq 1$
$$
\begin{array}{l}
\text { and } x^{2}+y^{2}=1, y^{2}+z^{2}=1 \\
\Rightarrow x^{2}+z^{2}=1 \\
\text { But } x^{2}+y^{2}=1 \Rightarrow y^{2}+x^{2}=1
\end{array}
$$
$$
\begin{array}{l}
\text { and } x^{2}+y^{2}=1, y^{2}+z^{2}=1 \\
\Rightarrow x^{2}+z^{2}=1 \\
\text { But } x^{2}+y^{2}=1 \Rightarrow y^{2}+x^{2}=1
\end{array}
$$
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