如果直線y=kx+1與圓x2+y2+kx+my-4=0交于M、N兩點(diǎn),且M、N關(guān)于直線x+y-1=0對(duì)稱,則k-m的值為________.
4
分析:因?yàn)橹本y=kx+1與圓x
2+y
2+kx+my-4=0的兩個(gè)交點(diǎn)關(guān)于直線x+y-1=0對(duì)稱,所以直線y=kx+1與直線x+y-1=0垂直,且直線x+y-1=0過圓x
2+y
2+kx+my-4=0的圓心.這樣直線y=kx+1與直線x+y-1=0垂直,斜率等于直線x+y-1=0的負(fù)倒數(shù),直線x+y-1=0過圓x
2+y
2+kx+my-4=0的圓心,則圓心坐標(biāo)滿足直線方程,就可求出k,m的值,解出k-m.
解答:∵直線y=kx+1與圓x
2+y
2+kx+my-4=0交于M、N兩點(diǎn),且M、N關(guān)于直線x+y-1=0對(duì)稱,
∴直線y=kx+1與直線x+y-1=0垂直,且直線x+y-1=0過圓x
2+y
2+kx+my-4=0的圓心.
∴k=1,
解得,m=-3
∴k-m=1-(-3)=4
故答案為4
點(diǎn)評(píng):本題主要考查直線與圓的位置關(guān)系的判斷,圓上兩點(diǎn)一定關(guān)于直徑所在的直線對(duì)稱.