题目内容
(2006•辽宁)如图,已知A(-1,0),E(0,-![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_ST/0.png)
(1)求证:直线FC是⊙A的切线;
(2)求点C的坐标及直线FC的解析式;
(3)有一个半径与⊙A的半径相等,且圆心在x轴上运动的⊙P.若⊙P与直线FC相交于M,N两点,是否存在这样的点P,使△PMN是直角三角形?若存在,求出点P的坐标;若不存在,请说明理由.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_ST/images1.png)
【答案】分析:(1)连接AF,由于AE∥BF,故∠1=∠3,∠4=∠2,又∵AB=AF,∴∠3=∠4∴∠1=∠2又∵AO=AF,AE=AE
∴△AOE≌△AFE∴∠AFE=∠AOE=90°∴FC是⊙O的切线.
(2)方法由(1)知EF=OE=
∵AE∥BF,∴
=
,∴
=
∴CE=
CO+
①(6分)∵OE2+OC2=CE2,∴CE2=(
)2+CO2②(7分)由①②解得OC=0(舍去)或OC=2,∴C(2,0)(8分)∵直线FC经过E(0,-
),C(2,0)两点,∴直线FC的解析式为y=
x-
.
解答:(1)证明:连接AF,
∵AE∥BF,
∴∠1=∠3,∠4=∠2,
又∵AB=AF,
∴∠3=∠4,
∴∠1=∠2,
又∵AO=AF,AE=AE,
∴△AOE≌△AFE,
∴∠AFE=∠AOE=90°,
∴FC是⊙O的切线.
(2)解:方法①由(1)知EF=OE=
,
∵AE∥BF,
∴
=
,
∴
=
,
∴CE=
CO+
①;
又∵OE2+OC2=CE2,
∴CE2=(
)2+CO2②;
由①②解得OC=0(舍去)或OC=2,
∴C(2,0),
∵直线FC经过E(0,-
),C(2,0)两点,
设FC的解析式:y=kx+b,
∴
,
解得
,
∴直线FC的解析式为y=
x-
.
方法②:
∵CF切⊙A于点F,
∴∠AFC=∠EOC=90°,
又∠ACF=∠OCE,
∴△COE∽△CFA,
∴
,
∴
,
即CE=
CO-
①;
又OE2+OC2=CE2,
∴CE2=(
)2+CO2②;
由①②解得CO=0(舍去)或CO=2;
∴C(2,0)
(求FC的解析式同上).
方法③∵AE∥BF,
∴
=
,
∴
=
,
∴CE=
CO+
①,
∵FC切⊙A于点F,
∴∠AFC=∠COE=90°,
∴∠ACE=∠OCE,
∴△COE∽△CFA,
∴
=
,
∴
=
,
∴CE=
CO-
②.
由①②解得:CO=2,
∴C(2,0),
(求FC的解析式同上).
(3)解:存在:
当点P在点C左侧时,若∠MPN=90°,过点P作PE⊥MN于点E,
∵∠MPN=90°,PM=PN,
∴PE=PM×cos45°=
,
∵AF⊥FC,
∴PE∥AF,
∴△CPE∽△CAF,
∴
=
,
∴
=
,
∴CP=
,
∴PO=
-2,
∴P(2-
,0).
当点P在点C右侧P′时,设∠M′P′N′=90°,过点P′作P′Q⊥M′N′于点Q,则P′Q=
.
∴P′Q=PE,可知P′与P关于点C中心对称,根据对称性得:
∴OP′=OC+CP′=2+
,
∴P′(2+
,0),
∴存在这样的点P,使得△PMN为直角三角形,P点坐标(2-
,0)或(2+
,0).
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/images54.png)
点评:本题是一道综合性很强的传统型压轴题,其难度比较恰当,选拔功能较强,解第3小题时要注意分类讨论,这是本题最容易失分的地方.
∴△AOE≌△AFE∴∠AFE=∠AOE=90°∴FC是⊙O的切线.
(2)方法由(1)知EF=OE=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/6.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/7.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/8.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/9.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/10.png)
解答:(1)证明:连接AF,
∵AE∥BF,
∴∠1=∠3,∠4=∠2,
又∵AB=AF,
∴∠3=∠4,
∴∠1=∠2,
又∵AO=AF,AE=AE,
∴△AOE≌△AFE,
∴∠AFE=∠AOE=90°,
∴FC是⊙O的切线.
(2)解:方法①由(1)知EF=OE=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/11.png)
∵AE∥BF,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/12.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/13.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/14.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/15.png)
∴CE=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/16.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/17.png)
又∵OE2+OC2=CE2,
∴CE2=(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/18.png)
由①②解得OC=0(舍去)或OC=2,
∴C(2,0),
∵直线FC经过E(0,-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/19.png)
设FC的解析式:y=kx+b,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/20.png)
解得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/21.png)
∴直线FC的解析式为y=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/22.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/23.png)
方法②:
∵CF切⊙A于点F,
∴∠AFC=∠EOC=90°,
又∠ACF=∠OCE,
∴△COE∽△CFA,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/24.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/25.png)
即CE=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/26.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/27.png)
又OE2+OC2=CE2,
∴CE2=(
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/28.png)
由①②解得CO=0(舍去)或CO=2;
∴C(2,0)
(求FC的解析式同上).
方法③∵AE∥BF,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/29.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/30.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/31.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/32.png)
∴CE=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/33.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/34.png)
∵FC切⊙A于点F,
∴∠AFC=∠COE=90°,
∴∠ACE=∠OCE,
∴△COE∽△CFA,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/35.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/36.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/37.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/38.png)
∴CE=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/39.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/40.png)
由①②解得:CO=2,
∴C(2,0),
(求FC的解析式同上).
(3)解:存在:
当点P在点C左侧时,若∠MPN=90°,过点P作PE⊥MN于点E,
∵∠MPN=90°,PM=PN,
∴PE=PM×cos45°=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/41.png)
∵AF⊥FC,
∴PE∥AF,
∴△CPE∽△CAF,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/42.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/43.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/44.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/45.png)
∴CP=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/46.png)
∴PO=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/47.png)
∴P(2-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/48.png)
当点P在点C右侧P′时,设∠M′P′N′=90°,过点P′作P′Q⊥M′N′于点Q,则P′Q=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/49.png)
∴P′Q=PE,可知P′与P关于点C中心对称,根据对称性得:
∴OP′=OC+CP′=2+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/50.png)
∴P′(2+
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/51.png)
∴存在这样的点P,使得△PMN为直角三角形,P点坐标(2-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/52.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/53.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650014_DA/images54.png)
点评:本题是一道综合性很强的传统型压轴题,其难度比较恰当,选拔功能较强,解第3小题时要注意分类讨论,这是本题最容易失分的地方.
![](http://thumb2018.1010pic.com/images/loading.gif)
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