If x,y∈R satisfies (x+5)2+(y−12)2=(14)2 then the minimum value of x2+y2 is
Let x+5=14cosθ and y−12=14sinθ
∴ x2+y2=(14cosθ−5)2+(14sinθ+12)2=196+25+144+28(12sinθ−5cosθ)=365+28(12sinθ−5cosθ)∴ x2+y2min=365−28×13=365−364=1