【答案】
分析:(1)①當直線經(jīng)過圓心M(4,2)時,將圓心坐標代入直線解析式,即可求得b的值;
②當若直線與⊙M相切,如答圖1所示,應有兩條符合條件的切線,不要遺漏.
欲求此時b的值,可以先求出切點P的坐標,代入解析式即可;欲求切點P的坐標,可以構(gòu)造相似三角形△PMN∽△BAO,求得PN=2MN,然后在Rt△PMN中利用勾股定理求出MN和PN,最后求出P點坐標;
(2)本問關鍵是弄清直線掃過矩形ABCD的運動過程,可以分為五個階段,分別求出每一階段S的表達式,如答圖2-4所示.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/images0.png)
解:(1)①直線l:y=-2x+b(b≥0)經(jīng)過圓心M(4,2)時,則有:2=-2×4+b,∴b=10;
②若直線l:y=-2x+b(b≥0)與⊙M相切,如答圖1所示,應有兩條符合條件的切線.
設直線與x軸、y軸交于A、B點,則A(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/0.png)
,0)、B(0,b),∴OB=2OA.
由題意,可知⊙M與x軸相切,設切點為D,連接MD;
設直線與⊙M的一個切點為P,連接MP并延長交x軸于點G;
過P點作PN⊥MD于點N,PH⊥x軸于點H.
易證△PMN∽△BAO,
∴PN:MN=OB:OA=2:1,
∴PN=2MN.
在Rt△PMN中,由勾股定理得:PM
2=PN
2+MN
2,解得:MN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/1.png)
,PN=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/2.png)
,
∴PH=ND=MD-MN=2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/3.png)
,OH=OD-HD=OD-PN=4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/4.png)
,
∴P(4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/5.png)
,2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/6.png)
),代入直線解析式求得:b=10-2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/7.png)
;
同理,當切線位于另外一側(cè)時,可求得:b=10+2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/8.png)
.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/images10.png)
(2)由題意,可知矩形ABCD頂點D的坐標為(2,2).
由一次函數(shù)的性質(zhì)可知,當b由小到大變化時,直線l:y=-2x+b(b≥0)向右平移,依次掃過矩形ABCD的不同部分.
可得當直線經(jīng)過A(2,0)時,b=4;當直線經(jīng)過D(2,2)時,b=6;當直線經(jīng)過B(6,0)時,b=12;當直線經(jīng)過C(6,2)時,b=14.
①當0≤b≤4時,S=0;
②當4<b≤6時,如答圖2所示.
設直線l:y=-2x+b與x軸交于點P,與AD交于點Q.
令y=0,可得x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/9.png)
,∴AP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/10.png)
-2;
令x=2,可得y=b-4,∴AQ=b-4.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/images13.png)
∴S=S
△APQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/11.png)
AP•AQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/12.png)
(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/13.png)
-2)(b-4)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/14.png)
b
2-2b+4;
③當6<b≤12時,如答圖3所示.
設直線l:y=-2x+b與x軸交于點P,與CD交于點Q.
令y=0,可得x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/15.png)
,∴AP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/16.png)
-2;
令y=2,可得x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/17.png)
-1,∴DQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/18.png)
-3.
S=S
梯形APQD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/19.png)
(DQ+AP)•AD=b-5;
④當12<b≤14時,如答圖4所示.
設直線l:y=-2x+b與BC交于點P,與CD交于點Q.
令x=6,可得y=b-12,∴BP=b-12,CP=14-b;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/images23.png)
令y=2,可得x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/20.png)
-1,∴DQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/21.png)
-3,CQ=7-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/22.png)
.
S=S
矩形ABCD-S
△PQC=8-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/23.png)
CP•CQ=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/24.png)
b
2+7b-41;
⑤當b>14時,S=S
矩形ABCD=8.
綜上所述,當b由小到大變化時,S與b的函數(shù)關系式為:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185421422178916/SYS201311011854214221789022_DA/25.png)
.
點評:本題是動線型壓軸題,綜合考查了一次函數(shù)的圖象與性質(zhì)、圓的切線性質(zhì)、相似三角形、矩形、梯形、勾股定理以及圖形面積等重要知識點,涉及的考點較多,難度較大,對同學們的解題能力提出了很高的要求.本題的難點在于:(I)第(1)②問中,圓的切線有兩條,容易遺漏.求切點坐標時候,注意運用相似關系化簡運算;(II)第(2)問中,動直線的運動過程分析是難點,注意劃分為五個階段,分別求出每個階段S的表達式.