问题164设点A(2,0),点B在圆x2+y2=1上,点C是∠AOB的角平分线与线段AB的交点,求当B运动时点C的轨迹方程.图1解答1如图1,设C(x,y),B(x1,y1).由于C是∠AOB的角平分线与线段AB的交点,根据角平分线定理有:|AC||BC|=|OA||OB|=2∴C为AB的定比分点,且λ=2由定比分点公式有:x=2+2x13y=2y Question 164 sets point A(2,0), point B is on the circle x2+y2=1, point C is the intersection point of the angle bisector of ∠AOB and line segment AB, and the trajectory equation of point C is obtained when B moves. Solution 1 As shown in Fig. 1, let C(x,y) and B(x1,y1). Since C is the intersection of the angle bisector of the AOB and the segment AB, the theorem of the angle bisector is: |AC||BC|= |OA||OB|=2∴C is the point where AB is scored, and λ=2 is determined by the point formula: x=2+2x13y=2y