In $\mathbb{R^2}$ we consider the relation
\begin{equation} (x,y)R(a,b)\Longleftrightarrow (x=a \,\text{and} \, y=b)\, \text{or}\,(x^2+y^2<a^2+b^2) \end{equation}
$R$ is an order relation if it is:
1) Reflexive: $\forall (x,y)\in \mathbb{R^2}, (x,y)R(x,y)$:
It's true because $x=x$ and $y=y$
2) Antisymmetric $(x,y)R(a,b) $ and $ (a,b)R(x,y)\Rightarrow \, (x,y)=(a,b)$
It's true because if $x^2+y^2<a^2+b^2$ and $a^2+b^2<x^2+y^2$ then $x=a$ and $y=b$
3) Transitive $(x,y)R(a,b) $ and $(a,b)R(c,d) $
It's true because: $x^2+y^2<a^2+b^2 $ and $a^2+b^2<c^2+d^2$ then $x^2+y^2<c^2+d^2$
So $R$ is an order relation.
Is correct my work?