在直角坐标系xoy中，圆O的方程为x2+y2=1．（1）若直线l与圆O切于第一

问题解答题

在直角坐标系xoy中，圆O的方程为x²+y²=1．

（1）若直线l与圆O切于第一象限且与坐标轴交于点A，B，当|AB|最小时，求直线l的方程；

（2）若A，B是圆O与x轴的交点，C是圆在直径AB的上方的任意一点，过该点作CD⊥AB交圆O于点D，当点C在圆O上移动时，求证：∠OCD的角平分线经过圆O上的一个定点，并求出该定点的坐标．

答案

（1）设直线l的方程为

=1(a＞0，b＞0)，则

=1，

∴

=1，∴ab≥2（当且仅当a=b=

时，取等号）

∴|AB|=

a2+b2

≥

2ab

≥2（当且仅当a=b=

时，取等号）

即|AB|最小为2，此时直线l的方程为x+y-

=0；

（2）证明：设∠OCD的角平分线为CP，交圆于P，则∠OCP=∠DCP

因为OC、OP为圆的半径，所以∠OCP=∠OPC，所以∠DCP=∠OPC

所以CD∥OP

因为CD⊥AB，A、B为定点，所以OP⊥AB

所以P为定点，坐标为（0，-1）

完形填空。
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