判定圆(x+6)^2+(y-2)^2=5与下列直线的位置关系:(1)2x+y+5=0(2)x+y+2=0(3)3x
人气:438 ℃ 时间:2020-05-09 23:01:14
解答
解
圆(x+6)^2+(y-2)^2=5的圆心是(-6,2)半径是√5
圆心到2x+y+5=0得距离是
|-6*2+2+5|/√(2²+1²)=√5=√5 所以与圆相切
圆心到x+y+2=0得距离是
|-6+2+2|√(1²+1²)=√2
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