已知圆(x-1)^2+(y+2)^2=4和直线2x+y-m=0相交于M、N两点,且角MON=120度,则m=?

已知圆(x-1)^2+(y+2)^2=4和直线2x+y-m=0相交于M、N两点,且角MON=120度,则m=?
已知圆(x-1)^2+(y+2)^2=4和直线2x+y-m=0相交于M、N两点,且角MON=120度,则m=?

已知圆(x-1)^2+(y+2)^2=4和直线2x+y-m=0相交于M、N两点,且角MON=120度,则m=?
O点是圆心
∠MON=120°
∵MO=ON
∴∠NMO=∠MNO
∴∠MNO=30°
则O点到直线的距离为半径的一半,∠MON=120°
(x-1)^2+(y+2)^2=4
O:(1,-2),r=2
|2*1-2-m|/√(2²+1²)=2*0.5
m=±√5

我也是南外的啊......
刚做完本想给下解答的，不过一打才发现实在是太烦了......
so。。。

同求~

把直线y=-2x+m代入圆(x-1)²+(y+2)²=4得，5x²-(10+4m)x+m²+4m+1=0,△=(10+4m)²-4×5(m²+4m+1)>0,
(5+2m)²-5(m²+4m+1)>0,20-m²>0,m²<20,
设M(x1,y1),N(x2,y2),则y1=-2x...

全部展开

把直线y=-2x+m代入圆(x-1)²+(y+2)²=4得，5x²-(10+4m)x+m²+4m+1=0,△=(10+4m)²-4×5(m²+4m+1)>0,
(5+2m)²-5(m²+4m+1)>0,20-m²>0,m²<20,
设M(x1,y1),N(x2,y2),则y1=-2x1+m,y2=-2x2+m,x1+x2=(10+4m)/5,x1x2=(m²+4m+1)/5,
则向量OM=(x1,y1),ON=(x2,y2),
OM·ON=x1x2+y1y2=-2(x1²+x2²)+m(x1+x2)=-2(x1+x2)²+4x1x2+m(x1+x2)
=-8(5+2m)²/25+4(m²+4m+1)/5+2m(5+2m)/5
=-8(5+2m)²/25+5(8m²+26m+4)/25=(8m²-30m-180)/25,
|OM|=√(x1²+y1²)=√(5x1²-4mx1+m²),
|ON|=√(x2²+y2²)=√(5x2²-4mx2+m²),
|OM|·|ON|=√(5x1²-4mx1+m²)√(5x2²-4mx2+m²)
=√{m^4+[5(x1²+x2²)-4m(x1+x2)]m²+(5x1²-4mx1)(5x2²-4mx2)}
=√{m^4+[5(x1+x2)²-20x1x2-4m(x1+x2)]m²+25(x1x2)²+16m²x1x2-20mx1x2(x1+x2)}
=√{m^4+[5(x1+x2)²-4x1x2-4m(x1+x2)]m²+25(x1x2)²-20mx1x2(x1+x2)}
=√{m^4+[4(5+2m)²/5-4(m²+4m+1)/5-8m(5+2m)/5]m²+(m²+4m+1)²-8m(5+2m)(m²+4m+1)/5}
=√{m^4+(-4m²+24m+96)m²/5+(m²+4m+1)²-8m(5+2m)(m²+4m+1)/5}
=√(m^4-8m³+2m²+1)
-1/2=cos120°=cos∠MON=OM·ON/(|OM|·|ON|)
=[(8m²-30m-180)/25]/√(m^4-8m³+2m²+1)
即-2(8m²-30m-180)=25√(m^4-8m³+2m²+1)，
16(4m²-15m-90)²=25²(m^4-8m³+2m²+1)，
解出 m,带回检验且m²<20

收起