2. If $k$ is a given real number, such that the following system of equations about $a, b$
$$
\left\{\begin{array}{l}
a+3 b-1=0, \\
a^{2}+b^{2}-4 a-6 b+13-k=0
\end{array}\right.
$$

has real solutions, then the range of values for $k$ is ( ).
(A) $k \geqslant 10$
(B) $k \geqslant 12$
(C) $k \geqslant 15$
(D) $k \in \mathbf{R}$
2. A.

Notice that $a^{2}+b^{2}-4 a-6 b+13-k=0$, which is $(a-2)^{2}+(b-3)^{2}=k(k \geqslant 0)$.
Thus, the system of equations has real solutions
$\Leftrightarrow k \geqslant 0$ and $\sqrt{k}$ is greater than or equal to the distance from the center $(2,3)$ to the line $a+3 b-1=0$.
$$
\begin{array}{l}
\Leftrightarrow \sqrt{k} \geqslant \frac{|2+3 \times 3-1|}{\sqrt{1+3^{2}}}=\sqrt{10} \\
\Leftrightarrow k \geqslant 10 .
\end{array}
$$