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

The correct option is Symmetric.

Let us analyze the properties of the relation $P$ defined on the set of real numbers $\mathbb{R}$.
The relation is given by:
$P=\left\{\right(x,y)\mid x^2+y^2=1,x,y\in \mathbb{R}\}$

1. Reflexivity:
For a relation to be reflexive, every element $x\in \mathbb{R}$ must satisfy $(x,x)\in P$. This requires:
$x^2+x^2=1\Rightarrow 2x^2=1$
This equation does not hold true for all real numbers $x$ (for example, if $x=2$, then $2\left(2^2\right)=8\ne 1$). Thus, $P$ is not reflexive.

2. Symmetry:
For a relation to be symmetric, if $(x,y)\in P$, then $(y,x)\in P$ must also be true.
Let $(x,y)\in P$. By definition, this means:
$x^2+y^2=1$
Since addition of real numbers is commutative, we can rewrite this equation as:
$y^2+x^2=1$
This shows that $(y,x)\in P$. Therefore, the relation $P$ is symmetric.

3. Transitivity:
For a relation to be transitive, if $(x,y)\in P$ and $(y,z)\in P$, then $(x,z)\in P$ must be true.
Let us consider a counterexample: Let $x=1$, $y=0$, and $z=1$.
Here, $1^2+0^2=1$, so $(1,0)\in P$.
Also, $0^2+1^2=1$, so $(0,1)\in P$.
However, for $(x,z)=(1,1)$, we have:
$1^2+1^2=2\ne 1$
Thus, $(1,1)\notin P$, which means the relation is not transitive.

4. Anti-symmetry:
For a relation to be anti-symmetric, if $(x,y)\in P$ and $(y,x)\in P$, then $x=y$ must hold.
If we take $x=1$ and $y=0$, we have $(1,0)\in P$ and $(0,1)\in P$, but $1\ne 0$. Thus, the relation is not anti-symmetric.

Consequently, the relation $P$ is symmetric.