Question

已知x,y是实数,且x2+y2-4x-6y+12=0,求:x2+y2的最值

Answer 1

x2+y2-4x-6y+12=0

(x-2)^2+(y-3)^2=1

x^2+y^2为(x-2)^2+(y-3)^2=1上（x,y)到(0,0)的距离.则

(2,3)到(0,0)的距离=√13

x^2+y^2的最小值为：√13-1；最大值为:√13+1