If real number x and y satisfy $(x+5)^2 +(y-12)^2=14^2$ then find the minimum value of $x^2 +y^2$ Problem : 
If real number x and y satisfy $(x+5)^2 +(y-12)^2=14^2$ then find the minimum value of $x^2 +y^2$
Please suggest how to proceed on this question... I got this problem from  [1]: http://www.mathstudy.in/

As mathlove has already identified   the curve to be circle $\displaystyle (x+5)^2+(y-12)^2=14^2$
But I'm not sure how to finish  from where he has left of without calculus. 
Here is one of the ways:
Using Parametric equation, any point $P(x,y)$ on the circle can be represented as $\displaystyle (14\cos\phi-5, 14\sin\phi+12)$ where $0\le \phi<2\pi$
So, $\displaystyle  x^2+y^2=(14\cos\phi-5)^2+(14\sin\phi+12)^2$
$\displaystyle=14^2+5^2+12^2+28(12\sin\phi-5\cos\phi)$
Now  can you derive $$-\sqrt{a^2+b^2}\le a\sin\psi-b\cos\psi\le\sqrt{a^2+b^2}?$$
and complete the answer?
