Question

已知点p(x,y)在圆xˆ2＋yˆ2－6x－6y＋14＝0上,求xˆ2＋yˆ2＋

Answer 1

x²+y²-6x-6y+14＝0→(x-3)²+(y-3)²＝2²,

依圆的参数方程,点P可设为(3+2cosθ,3+2sinθ).

∴x²+y²+2x+3

＝(3+2cosθ)²+(3+2sinθ)²+2(3+2cosθ)+3

＝31+20[(4/5)cosθ+(3/5)sinθ]

＝31+20sin(θ+φ) (其中tanφ＝4/3)

所以,

sin(θ+φ)＝1时,(x²+y²+2x+3)|max＝51;

sin(θ+φ)＝-1时,(x²+y²+2x+3)|min＝11.