Question

已知圆C:x平方＋（y-1)平方＝5,直线L:mx-y+1-m=0(1)求证：对于任意m属于R,直线L与圆C总有两个交点

Answer 1

(1)

直线L:mx-y+1-m=0,y - 1 = m(x - 1,总过点D(1,1)

圆C的圆心C(0,1),半径r = √5

D,C的纵坐标相同,距离为横坐标之差1 < 半径r

D在圆内,直线L与圆C总有两个交点

(2)

设AB的中点为E,

|AB| = √7,|AE| = √7/2

|CE| = C与直线L的距离d = |-1 + 1 - m|/√(m² + 1) = |m|/√(m² + 1)

r² = CA² = CE² + AE²

5 = d² + 7/4

d² = 13/4 = m²/(m² + 1)

m²/(m² + 1) < 1,与13/4 = m²/(m² + 1)矛盾,题有问题

改后可仿照做.