✔ 最佳答案
Sol
P(x,y)為圓(x+4)^2+(y-6)^2=5上一點
(x-2t-2)^2+(y-t+1)^2 為P到點Q(2t+2,t-1)距離平方
圓(x+4)^2+(y-6)^2=5圓心(-4,6)到Q(2t+2,t-1)距離平方D^2
D^2=(2t+2+4)^2+(t-1-6)^2
=(2t+6)^2+(t-7)^2
=4t^2+24t+36+t^2-14t+49
=5t^2+10t+85
=5(t^2+2t)+85
=5(t^2+2t+1)+85-5
=5(t+1)^2+80
t=-1時D最小值=√80
So
√[(x-2t-2)^2+(y-t-1)^2] 最小值
=√80-√5
=4√5-√5
=3√5