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) $\sqrt{2}$
(B)
2.【Analysis and Solution】 $(x+5)^{2}+(y-12)^{2}=14^{2}$ is a circle with center at point $C(-5,12)$ and radius 14. Let $P$ be any point on the circle, then $|O P| \geqslant|C P|-|O C|=$ $14-13=1$

When points $C, O, P$ are collinear, the equality holds, so the minimum value of $P$ to point $O$ is 1, hence the answer is B.

Solution $2\left\{\begin{array}{l}x=-5+14 \cos \theta \\ y=12+14 \sin \theta\end{array}\right.$,
at this time $x^{2}+y^{2}=14^{2}+12^{2}+5^{2}+28(-5 \cos \theta+12 \sin \theta)$
$\geqslant 14^{2}+13^{2}-28 \sqrt{5^{2}+12^{2}}$
$=1$.