第五题的第二小问怎么做啊，谢谢！！

Question

如题所述

Answer 1

(1)显然MN的中垂线为x轴, C在MN的中垂线上, 故C为直线2x-y-6=0与x轴的交点(3, 0)
CM = r = √[(0-3)²+(-2-0)²]= √13
圆C的方程: (x - 3)² + y² = 13

(2)设过C的直线斜率为k, 方程为y-0 = k(x-3), y = k(x-3)
分别与l1, l2的方程联立可得A((3k-2)/(k-2), 4k/(k-2)), B(3(k-1)/(k+1), -6k/(k+1))
C(3, 0)为AB的中点:
3 = [(3k-2)/(k-2) + 3(k-1)/k+1)]/2 ①
0 = [4k/(k-2) -6k/(k+1)]/2= 0 ②
由①可得k = 8
由②可得k = 8或k= 0
两条件均须满足, k = 8
y = 8(x - 3)