第1个回答 2012-10-13
圆C: (x-1)^2 + (y + 2)^2 = 9, C(1, -2), r = 3
设l在y轴上的截距为c, 方程为y = x + c, x - y + c =0
C与l的距离d = |1 + 2 + c|/√2 = |c+3|/√2
以AB为直径的圆半径R = √(r^2 - d^2) = √[9 -(c+3)^2/2]
原点O在以AB为直径的圆内, 原点与l的距离d' = |0 -0+c|/√2 =|c|/√2
d' < R
|c|/√2 < √[9 -(c+3)^2/2]
c^2/2 < 9 - (c+3)^2/2
2c^2 + 6c -9 < 0
(-3 - 3√3)/2 < c < (-3 + 3√3)/2本回答被网友采纳