2 If real numbers $x, y$ satisfy $(x+5)^{2}+(y-12)^{2}=14^{2}$, then the minimum value of $x^{2}+y^{2}$ is ( ).
(A) 2
(B) 1
(C) $\sqrt{3}$
(D) $\sqrt{2}$
Let $x+5=14 \cos \theta, y-12=14 \sin \theta, \theta \in[0,2 \pi)$, then
$$
\begin{aligned}
x^{2}+y^{2} & =(14 \cos \theta-5)^{2}+(14 \sin \theta+12)^{2} \\
& =14^{2}+5^{2}+12^{2}-140 \cos \theta+336 \sin \theta \\
& =365+28(12 \sin \theta-5 \cos \theta) \\
& =365+28 \times 13 \sin (\theta-\varphi) \\
& =365+364 \sin (\theta-\varphi),
\end{aligned}
$$

where $\tan \varphi=\frac{5}{12}$.
Therefore, when $\theta=\frac{3 \pi}{2}+\arctan \frac{5}{12}$, i.e., $x=\frac{5}{13}, y=-\frac{12}{13}$, $x^{2}+$ $y^{2}$ reaches its minimum value of 1. Hence, the correct choice is (B).