题目内容
(本小题满分12分)
已知⊙
的圆心
,被
轴截得的弦长为
.
(Ⅰ)求圆
的方程;
(Ⅱ)若圆
与直线
交于
,
两点,且
,求
的值.
已知⊙
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003848876302.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003848891517.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003848907262.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003848923383.png)
(Ⅰ)求圆
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003848876302.png)
(Ⅱ)若圆
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003848876302.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003848969549.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003848969298.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003848985292.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849016508.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849032277.png)
(1)
(2)![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849063341.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849047788.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849063341.png)
试题分析:解:(Ⅰ)设⊙
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003848876302.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849094261.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849110676.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849125341.png)
所以⊙
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003848876302.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849047788.png)
(Ⅱ)设A
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849172517.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849188542.png)
联立
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240038492031174.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849219946.png)
由已知可得,判别式
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849235766.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240038492501009.png)
由于OA⊥OB,可得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849266638.png)
又
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849266712.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849281863.png)
所以
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240038493131016.png)
解得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849328343.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849359407.png)
所以
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824003849063341.png)
点评:解决该试题的关键是根据圆心和半径的关系式来得到圆的方程,同时能联立方程组,求解相交点的坐标关系式,结合垂直关系,运用向量的数量积为零来得到参数的方程,求解得到结论,属于中档题。
![](http://thumb.zyjl.cn/images/loading.gif)
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