已知關(guān)于x,y的方程,x²+y²-y+m=0,若圓C與直線x+2y-4=0相交與M、N兩點(diǎn),且MN=√(4/5),求m的值?
x^2+y^2-y+m=0
x+2y-4=0 x=4-2y
代入圓:16-16y+4y^2+y^2-y+m=0 5y^2-17y+16+m=0
M(x1,y1) N(x2,y2)
x1=4-2y1
x2=4-2y2
y1+y2=17/5
y1y2=(16+m)/5
兩點(diǎn)距離的平方=(x2-x1)^2+(y2-y1)^2=4/5
(2y1-2y2)^2+(y2-y1)^2=4/5
(y2-y1)^2=4/25=(y1+y2)^2-4y1y2=289/25-(64+4m)/5
64+4m=285/5=57
m=-7/4