If $\mathrm{x},\mathrm{y}\in \mathrm{R}$ satisfies (x+ 5)² + (y-12)² - (14)²,  then the minimum value of $\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}$ is _______. 

Let $\mathrm{x}+5=14\cos \mathrm{\theta } \mathrm{and} \mathrm{y}-12=14\sin \mathrm{\theta }$
$\begin{array}{l}\therefore {\mathrm{x}}^2+{\mathrm{y}}^2=(14\cos \mathrm{\theta }-5{)}^2+(14\sin \mathrm{\theta }+12{)}^2\\ =196+25+144+28(12\sin \mathrm{\theta }-5\cos \mathrm{\theta })\\ =365+28(12\sin \mathrm{\theta }-5\cos \mathrm{\theta })\\ {\left.\therefore \sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}\right|}_{min}=\sqrt{365-28\times 13}\\ =\sqrt{365-364}\\ =1\end{array}$