已知定点A(-1,0),B(1,0)和圆(X-3)^2+(Y-4)^2=4上的动点P,求是|PA|^2+|PB|^2最小时点P的坐标

0sinb/cosb=3/4
cosb=4/5
3cosa+4sina=[(3/4)*cosa+sina]/4=(5/16)*sin(a+b)
1≥sin(a+b)≥-1
5/16≥3cosa+4sina≥-5/16
圆C:(x-3)^2+(y-4)^2=4
C(3,4),r=2
点P是圆C上的动点
x-3=r*cosa,x=3+2cosa
y-4=r*sina,y=4+2sina
(x-3)^2+(y-4)^2=4
x^2+y^2
=6x+8y-21
=6*(3+2cosa)+8(4+2sina)-21
=29+12cosa+16sina
P(x,y)
|PA|^2=(x+1)^2+y^2
|PB|^2=(x-1)^2+y^2
|PA|^2+|PB|^2
=2(x^2+y^2)+2
=2(29+12cosa+16sina)+2
=60+8(3cosa+4sina)
|PA|^2+|PB|^2最小=60-8*(5/16)=57.5
sin(a+b)=-1,a+b=270°,0180°sina3cosa+4sina=-5/16.(1)
(sina)^2+(cosa)^2=1.(2)
解上方程组(1)、(2),得
sina=,cosa=
x=3+2cosa
y=4+2sina
请自己解