Can we almost cover any shape in the plane by disjoint/tangent disks of prescribed radii? Let $(a_n)_{n \in \mathbb{Z}}$ be some given sequence of positive numbers, such that $\lim_{n \to -\infty}  a_n=0,\lim_{n \to \infty}  a_n=+\infty$.
Let $\Omega \subseteq \mathbb{R}^2$ be a bounded, connected, open subset, with Lipschitz boundary.
Let $\epsilon >0$. Does there exist a countable collection of closed disks $B(x_k,r_k)$ with the following properties:

*

*For each $k$, $B(x_k,r_k) \subseteq \Omega$ and $r_k \in \{a_n\}$.

*For every two distinct disks, either their intersection is empty, or they are tangent to each other. (The interiors of the disks do not intersect in any case).

*$\Omega \setminus\cup_{k} B(x_k,r_k)$ has Lebesgue measure less then $\epsilon$.


Here $B(x_k,r_k)$ denotes the closed Euclidean disk with radius $r_k$, centered around $x_k$.
It's possible that we only need the assumption $\lim_{n \to -\infty}  a_n=0$.
I am fine with assuming higher regularity of the boundary $\partial \Omega$, say $C^1$.
Comment:
I am not sure if this cane be done for any such sequence $a_n$. One could suppose e.g. that perhaps certain bounds on the ratios between the different $a_n$ are needed.
Perhaps we could start with an easier question: If there was no restriction on the radii at all, could such a disjoint\tangent cover be constructed?

This question resembles in a sense Apollonian gaskets.
 A: Here is an answer to the modified version of your question (when one does not make any assumptions on the radii):
Theorem. Let $\Omega$ be a domain (an open and connected subset) in the complex plane. Then there exists a collection of pairwise disjoint open round disks $B_k, k\in K$, in $\Omega$, such that the complement to their union in $\Omega$ has zero measure.
Proof. Even for this result, I do not know an easy proof.
Definition. A collection of pairwise disjoint open round disks $B_k$ in $\Omega$ is called a packing if each component of
$$
\Omega\setminus \bigcup_{k\in K} cl(B_k)
$$
is a triangle bounded by circular arcs.
Step 1. Suppose that $cl(\Omega)$ is compact and has piecewise-circular boundary. Then it admits a (possibly infinite) circle packing.
I do not know a reference for this, but it follows from the arguments in
Brooks, Robert, On the deformation theory of classical Schottky groups, Duke Math. J. 52, 1009-1024 (1985). ZBL0587.58060.
This construction proceed in several steps. The first one is to construct a finite partial packing of $\Omega$ so that each complementary component is either a circular triangle or a circular quadrilateral. Brooks then packs each complementary quadrilateral $Q_j$ so that (unless $Q_j$ admits a finite packing) the accumulation set of the packing of $Q_j$ is one of the vertices of $Q_j$.
Step 2. Suppose now that $\Omega$ is a general domain in the complex plane. Exhaust $\Omega$ by compact subsets $C_j$ with PL boundaries. (An exhaustion means that $C_{j-1}\subset int(C_j)$ for each $j$.) Each $\Omega_j= C_j \setminus cl(C_{j-1})$ is relatively compact in $\Omega$ with PL boundary. Hence, apply Step 1 to each $\Omega_j$ and obtain a packing ${\mathcal P}_j$ of it.
Step 3. Set ${\mathcal P}:=\bigcup_j {\mathcal P}_j$. This is a packing of $\Omega$. Now, apply the Apollonian packing construction to each complementary circular triangle of  ${\mathcal P}$. Since the measure of the residual set of the Apollonian packing is zero, and since there are only countably many of these triangles, we obtain the required collection of pairwise disjoint open disks $B_k$.  qed
My guess is that this proof can be streamlined as follows:
Conjecture. Every domain $\Omega$ in the complex plane admits a locally finite packing by round disks, where local finiteness means that each compact subset of $\Omega$ intersects only finitely many of these disks.

As for your original question, I think the answer is again positive for general domains (no Lipschitz condition is needed) and the proof hinges upon the case when your domain is a circular triangle with zero angles.
A: I think thus might be helpful:
Is it possible to calculate volume of a cube (with volume $L^3$) by filling it with small balls each with a radius $r_N$ and the balls are disjoint ? The answer is 'No'.
Let the number of balls be $N$
$$\lim\limits_{N\mapsto \infty} r_N=0$$
Is it always possibles to have $$\lim\limits_{N\mapsto \infty} N \frac {4\pi r_N^3}{3}=L^3$$ $\quad$
The problem is the same as filling balls of unit radius into cubes of larger and larger side length and taking the limit as the side length goes to $+\infty$. This converts the problem into the Kepler conjecture.
The theorem of Hales and co-authors, saying that the Kepler conjecture is true, shows that the ratio between the left and right sides of the conjectured equation  $$\lim\limits_{N\mapsto \infty} N \frac {4\pi r_N^3}{3}=L^3$$  is no greater than
$$\frac{\pi}{3 \sqrt{2}} = 0.740480489...
$$
