求过点A(1,-2)的所有直线被圆x2+y2=5截得线段中点的轨迹方程。

作者: tihaiku 人气: - 评论: 0
问题 求过点A(1,-2)的所有直线被圆x2+y2=5截得线段中点的轨迹方程。
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解析 占A存网卜.根据垂径定理可知.被圆截得线段中点的圆心0(0,0)连线必然垂直于直线AB,所以B点在以0A为直径的圆上 (盲角所对的弦为直径)。所以B在以为半径的圆上。故B点的轨迹方程为

相关内容:直线,线段,中点,轨迹,方程

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