For the function
$f(x, y) = \sqrt{(x-1)^2+(y-2)^2 -1}$
Would it be sufficient to write the domain as
$(x-1)^2 + (y-2)^2 ≤ 1$
or is that too vague? Then, the range would be $z ≥ 0$ but how could that be expressed in terms of $x$ if the value has to be greater than or equal to $1$?
Your idea is pretty fine, but be careful about the direction of the inequality. If you have $\sqrt{z}$, then you want $z \ge 0$, not $z \le 0$. Since $f : \Bbb R^2 \to \Bbb R$, you could write the domain by
$$\mathrm{dom}(f) = \left\{ (x,y) \in \Bbb R^2 \; \middle| (x-1)^2 + (y-2)^2 \ge 1 \right\}$$
as a result. The range of $f$ could simply be given by $\Bbb R_{\ge 0}$ if you want a symbol for it. You just want the points it is possible for $f$ to map to, regardless of the constraints that need be placed on $x,y$ in order for that to happen.
For instance, you could have $x=y-1$. Then the radicand is given by
$$(x-1)^2 + (y-2)^2 - 1 \longrightarrow 2(y-2)^2 - 1$$
Clearly the function $g(y) = \sqrt{2(y-2)^2 - 1}$, for wherever $y$ ensures $g(y)$ is defined, has range $\Bbb R_{\ge 0}$.