(x-3)^2+(y-3)^2=6
x-3=(√6)cost,x=3+(√6)cost
y-3=(√6)sint,y=3+(√6)sint(t∈R)
故x+y=3+(√6)cost+3+(√6)sint=6+2√3sin(t+π/4)∈[6-2√3,6+2√3]
故有:
(x+y)max=6+2√3
(x+y)min=6-2√3