For $(a,b),(c,d)\in\mathbb R\times\mathbb R$, define $(a,b)\mathrel\sim(c,d)$ if and only if $a^2+b^2=c^2+d^2$. $\sim$ is an equivalence relation on $\mathbb R\times\mathbb R$. Let $\mathbb R^*=\{x\in\mathbb R\mid x\geq0\}$.
I'd like some feedback on my proof of the following proposition:
Proposition. There is a one-to-one correspondence (bijection) between $\mathbb R^*$ and the set of all equivalence classes for this equivalence relation.
Proof. For any $(a,b)\in\mathbb R\times\mathbb R$, the equivalence class of $(a,b)$ is $[(a,b)]=\{(x,y)\in\mathbb R\times\mathbb R\mid x^2+y^2=a^2+b^2\}$. Let $E$ be the set of all equivalence classes for this equivalence relation. That is, $E=\{[(a,b)]\mid(a,b)\in\mathbb R\times\mathbb R\}$. Let $f:\mathbb R^*\to E$ be defined by $f(r)=\{(x,y)\in\mathbb R\times\mathbb R\mid x^2+y^2=r\}$ for all $r\in\mathbb R^*$. We will prove that $f$ is a bijection by showing that $f$ is injective and surjective.
Proof that $f$ is an injection: Let $r_1,r_2\in\mathbb R^*$ such that $r_1\neq r_2$. Then $f(r_1)=\{(x,y)\in\mathbb R\times\mathbb R\mid x^2+y^2=r_1\}$ and $f(r_2)=\{(x,y)\in\mathbb R\times\mathbb R\mid x^2+y^2=r_2\}$. $f(r_1)$ and $f(r_2)$ represent circles centred at the origin with radii $\sqrt{r_1}$ and $\sqrt{r_2}$ respectively. Their intersection is the intersection of these two circles. Observe that $$\begin{align*} f(r_1)\cap f(r_2)&=\{(x,y)\in\mathbb R\times\mathbb R\mid x^2+y^2=r_1=r_2\}&&\text{(the intersection of these two circles),}\\ &=\emptyset&&\text{(they don't intersect since $r_1\neq r_2$).} \end{align*}$$ Hence $f(r_1)\neq f(r_2)$.
Proof that $f$ is a surjection: Let $[(a,b)]\in E$ where $(a,b)\in\mathbb R\times\mathbb R$. As previously established, $[(a,b)]=\{(x,y)\in\mathbb R\times\mathbb R\mid x^2+y^2=a^2+b^2\}$. Observe that $(a^2+b^2)\in\mathbb R^*$ and $f(a^2+b^2)=\{(x,y)\in\mathbb R\times\mathbb R\mid x^2+y^2=a^2+b^2\}=[(a,b)]$. $$\tag*{$\blacksquare$}$$