问题
填空题
极坐标系下,直线ρcos(θ+
|
答案
将方程ρcos(θ+
)=1,π 4
即
ρcosθ-2 2
ρsinθ=1,化成直角坐标方程,2 2
x-y-
=0,2
圆ρ=
的直角坐标方程为:2
x2+y2=2,
圆心到直线的距离为
d=
=1<|-
|2 2
,2
故直线与圆相交,有两个交点.
故答案是:2.
极坐标系下,直线ρcos(θ+
|
将方程ρcos(θ+
)=1,π 4
即
ρcosθ-2 2
ρsinθ=1,化成直角坐标方程,2 2
x-y-
=0,2
圆ρ=
的直角坐标方程为:2
x2+y2=2,
圆心到直线的距离为
d=
=1<|-
|2 2
,2
故直线与圆相交,有两个交点.
故答案是:2.
阅读理解。 | |||||||||||
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