Question

（本小题满分12分）

如图，圆与轴相切于点，与轴正半轴相交于两点（点在点的左侧），且．

         （Ⅰ）求圆的方程；

         （Ⅱ）过点任作一条直线与椭圆相交于两点，连接，求证：．

 

Answer 1

【答案】

（Ⅰ）．（Ⅱ）见解析。

【解析】(I)由于圆与轴相切于点, 所以圆心坐标为,然后根据

建立关于r的方程求出r值，圆的标准确定.

(2)将y=0代入圆的方程求出M,N的坐标，然后再分两种情况证明.

(i) 当轴时，由椭圆对称性可知.

当与轴不垂直时，可设直线的方程为．证明,然后直线方程与椭圆方程联立借助韦达定理来解决即可.

（Ⅰ）设圆的半径为（），依题意，圆心坐标为．································ 1分

∵　

∴　，解得．····································································· 3分

∴　圆的方程为．······················································· 5分

（Ⅱ）把代入方程，解得，或，

即点，．····················································································· 6分

（1）当轴时，由椭圆对称性可知．······························· 7分

（2）当与轴不垂直时，可设直线的方程为．

联立方程，消去得，．······················ 8分

设直线交椭圆于两点，则

，．······································································· 9分

∵　，

∴　

．······························································· 10分

∵，

11分

∴　，．····························································· 12分

综上所述，．   13分