【答案】
分析:①
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104352026858720/SYS201311031043520268587023_DA/0.png)
表示圓(x-3)
2+y
2=4上的動點(diǎn)P(x,y)與原點(diǎn)連線的斜率,畫出滿足條件的圖象,分析后可得答案.
②先把y=0代入(x-1)
2+(y-b)
2=2求出對應(yīng)的x,即可求出被x軸截得的弦長,再結(jié)合已知條件即可求出b.
解答:解:①
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104352026858720/SYS201311031043520268587023_DA/1.png)
表示圓(x-3)
2+y
2=4上的動點(diǎn)P(x,y)與原點(diǎn)連線的斜率,
如下圖所示:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104352026858720/SYS201311031043520268587023_DA/images2.png)
設(shè)OP為y=kx,聯(lián)立(x-3)
2+y
2=4
得(k
2+1)x
2+-6x+5=0
令△=36-20(k
2+1)=0
解得k=±
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104352026858720/SYS201311031043520268587023_DA/2.png)
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104352026858720/SYS201311031043520268587023_DA/3.png)
的范圍是[-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104352026858720/SYS201311031043520268587023_DA/4.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104352026858720/SYS201311031043520268587023_DA/5.png)
]
②把y=0代入(x-1)
2+(y-b)
2=2得:
(x-1)
2+b
2=2⇒(x-1)
2=2-b
2⇒x1=1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104352026858720/SYS201311031043520268587023_DA/6.png)
,x2=1-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104352026858720/SYS201311031043520268587023_DA/7.png)
所以有:|x
1-x
2|=2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104352026858720/SYS201311031043520268587023_DA/8.png)
由題得:2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104352026858720/SYS201311031043520268587023_DA/9.png)
=2⇒
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104352026858720/SYS201311031043520268587023_DA/10.png)
=1⇒b=±1.
故答案為:[-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104352026858720/SYS201311031043520268587023_DA/11.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104352026858720/SYS201311031043520268587023_DA/12.png)
],±1.
點(diǎn)評:本題考查的知識點(diǎn)是直線與圓的位置關(guān)系,直線的斜率,其中第一空的關(guān)鍵是分析出
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103104352026858720/SYS201311031043520268587023_DA/13.png)
表示圓(x-3)
2+y
2=4上的動點(diǎn)P(x,y)與原點(diǎn)連線的斜率,第二空的關(guān)鍵是構(gòu)造關(guān)于b的方程.