题干

已知圆C:x2+y2﹣2x+4y﹣4=0,是否存在斜率为1的直线l,使l被圆C截得的弦长AB为直径的圆过原点,若存在求出直线的方程l,若不存在说明理由.

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答案(点此获取答案解析)

解:圆C化成标准方程为(x﹣1)2+(y+2)2=9,假设存在以AB为直径的圆M,圆心M的坐标为(a,b).

∵CM⊥l,即kCM•kl=b+2a-1

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