【答案】
分析:(1)因为直线l的解析式为y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/0.png)
x-3,并且与x轴、y轴分别相交于点A、B,所以分别令y=0;x=0,即可求出A、B的坐标;
(2)可设动圆的圆心在C处时与直线l相切,设切点为D,连接CD,则CD⊥AD,CD=1,由∠CAD=∠BAO,∠CDA=∠BOA=90°,可知Rt△ACD∽Rt△ABO,利用相似三角形对应边的比等于相似比,可得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/2.png)
,求出AC的值,即可得到此时OC的值,利用OC的长度结合速度即可求出时间;根据对称性,圆C还可能在直线l的右侧,与直线l相切,
此时OC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/3.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/4.png)
;
(3)可设在t秒时,动圆的圆心在F点处,动点在P处,此时OF=0.4t,BP=0.5t,F点的坐标为(0.4t,0),连接PF.
因为
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/5.png)
,又
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/6.png)
,所以可得到
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/7.png)
,进而可得到FP∥OB,PF⊥OA,所以P点的横坐标为0.4t,又结合P点在直线AB上,可得P点的纵坐标为0.3t-3,因此可见:当PF=1时,P点在动圆上,当0≤PF<1时,P点在动圆内,而当P=1时,由对称性可知,有两种情况:①当P点在x轴下方时,PF=-(0.3t-3)=1,解之可得t的值,②当P点在x轴上方时,PF=0.3t-3=1,解之得t的另一个值,进而可得到当
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/8.png)
时,0≤PF≤1,并且此时点P在动圆的圆面上,所经过的时间为
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/9.png)
.
解答:解:(1)在y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/10.png)
x-3中,令x=0,得y=-3;令y=0,得x=4,故得A、B两的坐标为
A(4,0),B(0,-3). (2分)
(2)若动圆的圆心在C处时与直线l相切,设切点为D,如图所示.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/images11.png)
连接CD,则CD⊥AD. (3分)
由∠CAD=∠BAO,∠CDA=∠BOA=90°,可知Rt△ACD∽Rt△ABO,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/11.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/12.png)
,则AC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/13.png)
. (4分)
此时OC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/14.png)
(秒). (5分)
根据对称性,圆C还可能在直线l的右侧,与直线l相切,
此时OC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/15.png)
. (7分)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/16.png)
(秒).
答:(略). (8分)
(3)设在t秒,动圆的圆心在F点处,动点在P处,此时OF=0.4t,BP=0.5t,F点的坐标为(0.4t,0),连接PF,
∵
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/17.png)
,又
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/18.png)
,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/images20.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/19.png)
,
∴FP∥OB,∴PF⊥OA(9分)
∴P点的横坐标为0.4t,
又∵P点在直线AB上,
∴P点的纵坐标为0.3t-3,
可见:当PF=1时,P点在动圆上,当0≤PF<1时,P点在动圆内. (10分)
当PF=1时,由对称性可知,有两种情况:
①当P点在x轴下方时,PF=-(0.3t-3)=1,解之得:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/20.png)
;
②当P点在x轴上方时,PF=0.3t-3=1,解之得:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/21.png)
. (11分)
∴当时
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/22.png)
时,0≤PF≤1,此时点P在动圆的圆面上,所经过的时间为
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/23.png)
,
答:动点在动圆的圆面上共经过了
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103201722232952968/SYS201311032017222329529024_DA/24.png)
秒. (12分)
点评:本题是一道综合性强的题目,解决这类问题常用到分类讨论、数形结合、方程和转化等数学思想方法.