【答案】
分析:(1)連接AC,由于BC與⊙A相切,則AC⊥BC,在Rt△ABC中,OC⊥AB,根據(jù)射影定理即可求得OC的長,從而得到C點(diǎn)的坐標(biāo),進(jìn)而用待定系數(shù)法求出直線BC的解析式.
(2)可設(shè)出G點(diǎn)的坐標(biāo)(設(shè)橫坐標(biāo),利用直線BC的解析式表示縱坐標(biāo)),連接AP、AG;由于GC、GP都是⊙A的切線,那么∠AGC=∠ABP=60°,在Rt△AGC中,AC的長易求得,根據(jù)∠AGC的度數(shù),即可求得AG的長;過G作GH⊥x軸于H,在Rt△GAH中,可根據(jù)G點(diǎn)的坐標(biāo)表示出AH、GH的長,進(jìn)而由勾股定理求得G點(diǎn)的坐標(biāo).
(3)若⊙A與直線交于點(diǎn)E、F,則AE=AF,如果△AEF是直角三角形,則∠EAF必為直角,那么△EAF是以A為頂點(diǎn)的等腰直角三角形,因此可分作兩種情況考慮:
①點(diǎn)A在B點(diǎn)右側(cè)時(shí),可過A作直線BC的垂線,設(shè)垂足為M,在(2)題已經(jīng)求得了⊙A的半徑,即可得到AM的長,易證得△BAM∽△BCO,通過相似三角形所得比例線段即可求得AB的長,進(jìn)而可得到OA的長,從而得出A點(diǎn)的坐標(biāo);
②點(diǎn)A在B點(diǎn)左側(cè)時(shí),方法同①.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/images0.png)
解:(1)如圖1所示,連接AC,則AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/0.png)
,
在Rt△AOC中,AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/1.png)
,OA=1,則OC=2,
∴點(diǎn)C的坐標(biāo)為(0,2);
設(shè)切線BC的解析式為y=kx+b,它過點(diǎn)C(0,2),B(-4,0),
則有
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/2.png)
,解之得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/3.png)
;
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/4.png)
.(4分)
(2)如圖1所示,設(shè)點(diǎn)G的坐標(biāo)為(a,c),過點(diǎn)G作GH⊥x軸,垂足為H點(diǎn),
則OH=a,GH=c=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/5.png)
a+2,(5分)
連接AP,AG;
因?yàn)锳C=AP,AG=AG,所以Rt△ACG≌Rt△APG(HL),
所以∠AGC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/6.png)
×120°=60°,
在Rt△ACG中,∠AGC=60°,AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/7.png)
,
∴sin60°=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/8.png)
,∴AG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/9.png)
;(6分)
在Rt△AGH中,AH=OH-OA=a-1,GH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/10.png)
a+2,
∵AH
2+GH
2=AG
2,
∴(a-1)
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/12.png)
,
解之得:a
1=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/13.png)
,a
2=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/14.png)
(舍去);(7分)
∴點(diǎn)G的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/15.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/16.png)
+2).(8分)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/images18.png)
(3)如圖2所示,在移動(dòng)過程中,存在點(diǎn)A,使△AEF為直角三角形.(9分)
要使△AEF為直角三角形,∵AE=AF,
∴∠AEF=∠AFE≠90°,∴只能是∠EAF=90°;
當(dāng)圓心A在點(diǎn)B的右側(cè)時(shí),過點(diǎn)A作AM⊥BC,垂足為點(diǎn)M,
在Rt△AEF中,AE=AF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/17.png)
,
則EF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/18.png)
,AM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/19.png)
EF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/20.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/21.png)
;
在Rt△OBC中,OC=2,OB=4,則BC=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/22.png)
,
∵∠BOC=∠BMA=90°,∠OBC=∠OBM,
∴△BOC∽△BMA,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/23.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/24.png)
,
∴AB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/25.png)
,
∴OA=OB-AB=4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/26.png)
,
∴點(diǎn)A的坐標(biāo)為(-4+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/27.png)
,0);(11分)
當(dāng)圓心A在點(diǎn)B的左側(cè)時(shí),設(shè)圓心為A′,過點(diǎn)A′作A′M′⊥BC于點(diǎn)M′,可得:
△A′M′B≌△AMB,A′B=AB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/28.png)
,
∴OA′=OB+A′B=4+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/29.png)
,
∴點(diǎn)A′的坐標(biāo)為(-4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/30.png)
,0);
綜上所述,點(diǎn)A的坐標(biāo)為(-4+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/31.png)
,0)或(-4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231152954209218/SYS201310212311529542092013_DA/32.png)
,0).(13分)
點(diǎn)評(píng):此題考查的知識(shí)點(diǎn)有:一次函數(shù)解析式的確定、勾股定理、切線的性質(zhì)、切線長定理、全等三角形及相似三角形的判定和性質(zhì)等;需要注意的是(3)題中,一定要考慮到點(diǎn)A在B點(diǎn)左側(cè)時(shí)的情況,以免漏解.