If (x, y) satisfies the equation (x+5)2+(y−12)2=196 then minimum value of x2+y2 is

Let  x+15=14cosθ, y-12=14sinθ ⇒(x+5)2+(y−12)2=(14)2Now         x=−5+14cosθand          y=12+14sinθ⇒          x2+y2=365+28(12sinθ−5cosθ)∴      min=365−28×13=1