If (x, y) satisfies the equation (x+5)2+(y−12)2=196 then minimum value of x2+y2 is
0
-1
1
-2
Let x+15=14cosθ, y-12=14sinθ
⇒(x+5)2+(y−12)2=(14)2
Now x=−5+14cosθ
and y=12+14sinθ
⇒ x2+y2=365+28(12sinθ−5cosθ)
∴ min=365−28×13=1