∵P(x,y)满足(x-3)2+(y-4)2=4, 则P点的坐标可表示为x=3+2cosa,y=4+2sina ||2+||2=[(4+2cosa)2+(4+2sina)2]+[(2+2cosa)2+(4+2sina)2] =[16+16cosa+4cos2a+16+16sina+4sin2a]+[4+8cosa+4cos2a+16+16sina+4sin2a] =(32+4+16cosa+16sina)+(20+4+8cosa+16sina) =60+24cosa+32sina =60+8(3cosa+4sina) =60+8[5sin(a+b)]>=60+8×(-5)=20 则||2+||2的最小值是20 故答案为:20 |