直线mx+ny-3=0和圆x^2+y^2+4x-1=0相切与点（-1,2）,求m,n的值.

把(-1,2)代入直线 -m+2n-3=0
m=2n-3
x^2+4x+y^2=1
x^2+4x+4+y^2=5
(x+2)^2+y^2=5
圆以(-2,0)为圆心,根号5为半径
相切圆心到直线距离为根号5
d=根号5=|-2m-3|/(根号(m^2+n^2))
5m^2+5n^2=4m^2+9+12m
m^2-12m+5n^2=9
4n^2+9-12n-24n+36+5n^2=9
9n^2-36n+36=0
n^2-4n+4=0
n=2
m=2n-3=1