  We consider the self-adjoint Landau Hamiltonian $H_0$ in $L^2(\mathbb{R}^2)$
whose spectrum consists of infinitely degenerate eigenvalues $\Lambda_q$, $q
\in \mathbb{Z}_+$, and the perturbed operator $H_\upsilon = H_0 +
\upsilon\delta_\Gamma$, where $\Gamma \subset \mathbb{R}^2$ is a regular Jordan
$C^{1,1}$-curve, and $\upsilon \in L^p(\Gamma;\mathbb{R})$, $p>1$, has a
constant sign. We investigate ${\rm Ker}(H_\upsilon -\Lambda_q)$, $q \in
\mathbb{Z}_+$, and show that generically $$0 \leq {\rm dim \, Ker}(H_\upsilon
-\Lambda_q) - {\rm dim \, Ker}(T_q(\upsilon \delta_\Gamma)) < \infty,$$ where
$T_q(\upsilon \delta_\Gamma) = p_q (\upsilon \delta_\Gamma)p_q$, is an operator
of Berezin-Toeplitz type, acting in $p_q L^2(\mathbb{R}^2)$, and $p_q$ is the
orthogonal projection on ${\rm Ker}\,(H_0 -\Lambda_q)$. If $\upsilon \neq 0$
and $q = 0$, we prove that ${\rm Ker}\,(T_0(\upsilon \delta_\Gamma)) = \{0\}$.
If $q \geq 1$, and $\Gamma = \mathcal{C}_r$ is a circle of radius $r$, we show
that ${\rm dim \, Ker} (T_q(\delta_{\mathcal{C}_r})) \leq q$, and the set of $r
\in (0,\infty)$ for which ${\rm dim \, Ker}(T_q(\delta_{\mathcal{C}_r})) \geq
1$, is infinite and discrete.
