题目内容
(2005•甘肃)如图,半圆O的直径AB=4,与半圆O内切的动圆O1与AB切于点M,设⊙O1的半径为y,AM=x,则y关于x的函数关系式是( )![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470024060/SYS201310191058594700240009_ST/images0.png)
A.y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470024060/SYS201310191058594700240009_ST/0.png)
B.y=-x2+
C.y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470024060/SYS201310191058594700240009_ST/1.png)
D.y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470024060/SYS201310191058594700240009_ST/2.png)
【答案】分析:连接01M,OO1,可得到直角三角形OO1M,在直角三角形中,利用勾股定理即可解得.
解答:
解:连接01M,OO1,可得到直角三角形OO1M,
依题意可知⊙O的半径为2,
则OO1=2-y,OM=2-x,O1M=y.
在Rt△OO1M中,由勾股定理得(2-y)2-(2-x)2=y2,
解得y=-
x2+x.
故选A.
点评:作连心线,连接圆心和切点得到直角三角形是常用的辅助线作法是本题的考查对象.
解答:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470024060/SYS201310191058594700240009_DA/images0.png)
依题意可知⊙O的半径为2,
则OO1=2-y,OM=2-x,O1M=y.
在Rt△OO1M中,由勾股定理得(2-y)2-(2-x)2=y2,
解得y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105859470024060/SYS201310191058594700240009_DA/0.png)
故选A.
点评:作连心线,连接圆心和切点得到直角三角形是常用的辅助线作法是本题的考查对象.
![](http://thumb2018.1010pic.com/images/loading.gif)
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