China Mathematical Competition

Let $x+5=14\cos \theta$ and $y-12=14\sin \theta$, for $\theta \in [0,2\pi )$. Hence $$\begin{array}{rl}{x}^2+y^2& =(14\cos \theta -5{)}^2+(14\sin \theta +12{)}^2\\ & =14^2+5^2+12^2-140\cos \theta +336\sin \theta \\ & =365+28(12\sin \theta -5\cos \theta )\end{array}$$ $$\begin{array}{rl}& =365+28\times 13\sin (\theta -\phi )\\ & =365+364\sin (\theta -\phi ),\end{array}$$ where $\tan \phi =\frac{5}{12}$. So $x^2+y^2$ has the minimum value $1$, when $\theta =\frac{3\pi }{2}+\arctan \frac{5}{12}$, i.e. $x=\frac{5}{13}$ and $y=-\frac{12}{13}$. Answer: B.