I have a family of sets $(B_r)_{r>0}$ defined by $$B_r=\{(x,y)\in \mathbb{R}^2, (x-3)^2+y^2<r^2\}.$$ Is $$\bigcup_{r>0}B_r =\mathbb{R}^2 $$
I try this : we always know that: $\bigcup B_r\subset \mathbb{R}^2$ but if we let $(a,b)\in \mathbb{R}^2$, we gat $(a-3)^2+b^2=r^2$ then $(a,b)\in B_{r+\varepsilon}\subset \bigcup B_r$
$B_r$ is a ball of radius $r$ around $(3,0)$.
So if you want it to include $(a,b)$ you need the radius $r$ to be larger than $d((a,b),(3,0)) = \sqrt{(a-3)^2+b^2}$.
where $d$ is the distance between two points in $\mathbb{R}^2$. $$d((x_1,y_1),(x_2,y_2))= \sqrt{(x_1-x_2)^2 +(y_1-y_2)^2}$$
So, if you choose $r=2\sqrt{(a-3)^2+b^2}$ then by definition $(a,b)\in B_r$.