Question

详细解释下此题，好吗，谢谢

Answer 1

圆心在2x+y=0上
圆心(m, -2m)
圆方程： (x-m)^2+(y+2m)^2 = r^2 (1)
过点(2,-1)
(2-m)^2+(-1+2m)^2 = r^2
5m^2-8m+5=r^2 (2)
from (1) and (2)
(x-m)^2+(y+2m)^2 = 5m^2-8m+5 (3)
直线: x-y-1=0 相切 (4)
sub (4) into (3)
(x-m)^2+(y+2m)^2 = 5m^2-8m+5
(x-m)^2+(1-x+2m)^2 = 5m^2-8m+5
2x^2 -(6m+2)x + 5m^2+4m+1 =5m^2-8m+5
2x^2 -(6m+2)x + 12m-4 =0
x^2 -(3m+1)x + 6m-2 =0
Δ=0
(3m+1)^2 -4(6m-2) =0
9m^2-18m+9=0
m^2-2m+1=0
m=1
r^2= (2-m)^2+(-1+2m)^2 = 1 + 1 =2
圆方程： (x-1)^2+(y+2)^2 = 2

追问

sub (4) into (3)
(x-m)^2+(y+2m)^2 = 5m^2-8m+5
(x-m)^2+(1-x+2m)^2 = 5m^2-8m+5
2x^2 -(6m+2)x + 5m^2+4m+1 =5m^2-8m+5
2x^2 -(6m+2)x + 12m-4 =0
x^2 -(3m+1)x + 6m-2 =0请问这些是什么意思？从哪里得的？