How to do partitions when they involve equations

Question

$B_r $={(x,y):$x^2 +y^2=$r}for each r $\in $ R

$ B_r$ is an index family of classes (Ref pg 40 A book of set theory.Pinter .electronic version 45 of224.or on google version 40 of 224

P1.∀r,s∈R, $B_r∩B_s=Ø$ Or $B_r=B_s$

1. I have to show each circle is disjoint for each $r$ $\in$ R

2. it’s union is B.

( This is either to obvious or involves more work .I am new to this subject. Working with sets and modulo ok)

Proof by contradiction

Assume they aren’t disjoint then (a,b) $\in $$ B_r $ $\cap $ $B_s$-> $a^2+b^2=r^2$ and $a^2+b^2=s^2$ Then$r^2=s^2$

These concentric circles at (0,0) share origin then r=s

So r=s & r!=s. Contradiction. So P1 is satisfied .

Since they are disjoint it’s union is B

Answer 1

Assume (a,b) in $B_r \cap B_r$.
Then r = $a^2 + b^2$ = s.
Take it from there.

Notice that B$_r$ is not a partition. Why?
What is B?