已知P(x,y)为圆C:(x+3)^2+(y-4)^2=1上任意一点,求x-2y最值

令x+3=cosa
x=-3+cosa
(y-4)²=1-cos²a=sin²a
y=4+sina
x-2y
=-3+cosa-8-2sina
=-(2sina-cosa)-11
=-√5sin(a-b)-11
其中tanb=1/2
所以最大值=√5-11,最小值-√5-11