China Mathematical Competition

Let $x+5 = 14\cos\theta$ and $y-12 = 14\sin\theta$, for $\theta \in [0, 2\pi)$.
Hence
$$\begin{align*} x^2 + y^2 &= (14\cos\theta - 5)^2 + (14\sin\theta + 12)^2 \\ &= 14^2 + 5^2 + 12^2 - 140\cos\theta + 336\sin\theta \\ &= 365 + 28(12\sin\theta - 5\cos\theta) \end{align*}$$
$$\begin{aligned} &=365 + 28 \times 13\sin(\theta - \varphi) \\ &=365 + 364\sin(\theta - \varphi), \end{aligned}$$
where $\tan \varphi = \frac{5}{12}$.
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.