直线与圆的方程求解题。
2个回答
展开全部
切线 x+y-1=0
圆心C在直线 :y=-2x 上
C(m, -2m)
A(2,-1)
C 到 切线 x+y-1=0 的距离
=r
= |m-2m-1|/√2
= |-m-1|/√2
|AC| = C 到 切线 x+y-1=0 的距离 =r
|AC|^2 =r^2
(m-2)^2+(-2m+1)^2 = (1/2)(m+1)^2
(m^2-4m+4)+(4m^2-4m+1) = (1/2)(m^2+2m +1)
5m^2-8m+5= (1/2)(m^2+2m +1)
10m^2-16m+10= m^2+2m +1
9m^2-18m+9=0
m^2-2m+1=0
m=1
C(1, -2)
r =|AC| =√[(m-2)^2+(-2m+1)^2 ]=√2
圆的方程
(x-1)^2+(y+2)^2 =2
圆心C在直线 :y=-2x 上
C(m, -2m)
A(2,-1)
C 到 切线 x+y-1=0 的距离
=r
= |m-2m-1|/√2
= |-m-1|/√2
|AC| = C 到 切线 x+y-1=0 的距离 =r
|AC|^2 =r^2
(m-2)^2+(-2m+1)^2 = (1/2)(m+1)^2
(m^2-4m+4)+(4m^2-4m+1) = (1/2)(m^2+2m +1)
5m^2-8m+5= (1/2)(m^2+2m +1)
10m^2-16m+10= m^2+2m +1
9m^2-18m+9=0
m^2-2m+1=0
m=1
C(1, -2)
r =|AC| =√[(m-2)^2+(-2m+1)^2 ]=√2
圆的方程
(x-1)^2+(y+2)^2 =2
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