Man, I'm having trouble with existence proofs.

How do I prove...

If $|a|<2r$, then there exists two points of intersection for the two circles whose equations are $ x^2+y^2=r^2$ and $(x-a)^2+y^2=r^2$?

I'm not sure how to express the two solutions backward. Also, the inequality is giving me grief.I want to know how to construct this proof from scratch. Not just use the given antecedent to justify the consequent.

I start by solving the system. But I have trouble. ...

$x^2+y^2=r^2\implies(x-a)^2=0$ Or $(\sqrt{r^2-x^2}-a)^2+\sqrt{r^2-x^2}^2=r^2\implies{}...$