若实数x,y满足x^2+y^2-6x-4y+12=0,则x^2+y^2的取值范围是

题目

若实数x,y满足x^2+y^2-6x-4y+12=0,则x^2+y^2的取值范围是

答案

x^2+y^2-6x-4y+12=(x-3)^2+(y-2)^2-1=0
(x-3)^2+(y-2)^2=1
x=3+cosa,y=2+sina
x^2+y^2=13+1+6cosa+4sina=14+2*13^(1/2)sin(a+@)
14-2*13^(1/2)