I have to obtain an union:
$\bigcup_{r\in R}\ \{(x,y): (x-r)^2 + (y+2r)^2 < r^2+1\}$
I know this is a series of circles that is limited by two hyperbolic functions that are symmetrical in relation to $y = -2x$. But how to get an union? I tried putting $y= \frac{1}{2}x$ into the circle's equation to get the intersection points but it works only for $r = 1$. It would require putting $y = \frac{1}{2}x + b, \forall b \in R$ but it doesn't look as a significant simplification. I have also thought of representing an intersection as a translation from $(x, -2x)$ by a vector $\pm\ [\frac{\sqrt{5}}{5} \sqrt{r^2+1}, \frac{2\sqrt{5}}{5}\sqrt{r^2+1}]$ but it hasn't turned out to be usefull aswell. I'm running out of ideas. I'd be really thankful for any kind of help.
You want the envelope of the family of circles given by varying $r$. The union will then be everything "inside" that envelope.
You can find the envelope of the family of curves $f(x,y,r) = 0$ by solving $$f(x,y,r) = \frac{\partial f}{\partial r}(x,y,r) = 0.$$ In this case (and generally), you should end up by determining the envelope parametrically as functions of $r$. When you eliminate $r$, you should find the equation $$3x^2+4xy=4,$$ which, as you suggested, is a hyperbola.