已知直角坐标系中,点P(x,y)满足x2−4+(y+3)2=0,则点P坐标为( ) A.(2,-3) B.(-2,3) C.(2,3) D.(2,-3)或(-2,-3)
问题描述:
已知直角坐标系中,点P(x,y)满足
+(y+3)2=0,则点P坐标为( )
x2−4
A. (2,-3)
B. (-2,3)
C. (2,3)
D. (2,-3)或(-2,-3)
答
∵
+(y+3)2=0,
x2−4
∴
,
x2−4=0 y+3=0
解得
或
x=2 y=−3
,
x=−2 y=−3
∴点P坐标为(2,-3)或(-2,-3).
故选:D.