${\text{If real numbers x and y satisfy (x+5)}}^2+{(y-12)}^2={14}^2, then the minimum value of \sqrt{x^2+y^2} is$

$$\begin{gathered} \text{Let x+5=14cosθ and y-12=14 sinθ} \\ {\text{x}}^2+y^2={(14cos\theta -5)}^2+{(14sin\theta +12)}^2 \\ =196+25+144+28(12sin\theta -5cos\theta ) \\ =365+28(12sin\theta -5cos\theta ) \\ \therefore {\sqrt{x^2+y^2}}_{min}=\sqrt{365-28x13} =\sqrt{365-364} =1 \end{gathered}$$