If real numbers  x and y satisfy (x+5)2+(y−12)2=(14)2 then the maximum value of x2+y2 is _________

Let x+5=14cos⁡θ and y−12=14sin⁡θ Thenx2+y2=(14cos⁡θ−5)2+(14sin⁡θ+12)2 =196+25+144+28(12sin⁡θ−5cos⁡θ) =365+28(12sin⁡θ−5cos⁡θ)∴ x2+y2max.=365+28×13=365+364=27