Question

已知圆C:x²+y²-2x+4y-4=0,问是否在斜率为1直线l,使以l被圆C截得的弦AB为直径的圆原

已知圆C:x²+y²-2x+4y-4=0,问是否在斜率为1直线l,使以l被圆C截得的弦AB为直径的圆原点,求出直线L的方程

Answer 1

x²+y²-2x+4y-4=0
(x - 1)² + (y + 2)² = 9
C(1, -2), 半径R = 3

设AB的中点为D, 新圆半径为r
CD斜率为 -1, 方程为: y + 2 = -(x - 1), y = -x -1

直线l: y = x + b, x - y + b = 0
交点D(-(b + 1)/2, (b - 1)/2)
AD = r = OD, r² = (b + 1)²/4 + (b - 1)²/4

AC² = AD² + CD² = r² + CD²

9 = (b + 1)²/4 + (b - 1)²/4 + [1 + (b + 1)/2]² + [-2 - (b -1)/2]²
(b + 4)(b - 1) = 0
b = -4, y = x - 4

或b = 1, y = x + 1