Question

38题第二小问怎么写

38题第二小问怎么写跪求数学大神

Answer 1

圆形: (x-2)^2+(y-1)^2 = 25 (1)
直线 :
y+2 = m(x-1)
mx-y -(m+2)= 0
圆心(2,1) 到 直线 : mx-y -(m+2)= 0 的距离
d=|2m - 1- (m+2) | /√(m^2 +1)
=| m-3| /√(m^2 +1)
d^2 + (2√5)^2 =r^2
(m-3)^2/(m^2 +1) + 20 =25
(m-3)^2 = 5(m^2 +1)
4m^2+6m-4=0
2m^2+3m-2=0
(2m-1)(m+2)=0
m=1/2 or -2
直线 : mx-y -(m+2)= 0
(1/2)x-y -(1/2+2)= 0 or -2x-y -(-2+2)= 0
(1/2)x -y - 5/2 =0 or -2x-y=0
x-2y-5=0 or 2x+y=0

追问

写的很好，但我看不懂