题目内容
(2007•衡阳)如图,点P在y轴上,⊙P交x轴于A,B两点,连接AP并延长交⊙P于C点,过点C的直线y=-2x+b交x轴于点D,交y轴于点E,且⊙P的半径为![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231655335248962/SYS201310212316553352489003_ST/0.png)
(1)求点P,点C的坐标;
(2)求证:CD是⊙P的切线;
(3)若二次函数y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231655335248962/SYS201310212316553352489003_ST/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231655335248962/SYS201310212316553352489003_ST/images2.png)
【答案】分析:(1)连接CB,根据已知及勾股定理等即可求解;
(2)只要证明∠ACD=90°即可得到DC是⊙P的切线.
(3)把A,C两点代入解析式求出未知数的值,进而求出其解析式;可求二次函数y=-
x2+
x+3与一次函数y=-2x+6的交点C和D,由此可知,满足条件的x的取值范围.
解答:
(1)解:如图,连接CB,
∵OP⊥AB,
∴OB=OA=2.(1分)
∵OP2+AO2=AP2
∴OP2=5-4=1,OP=1,(2分)
∵AC是⊙P的直径,
∴∠ABC=90°.
∵CP=PA,BO=OA,
∴BC=2PO=2.
∴P(0,1),C(2,2).(3分)
(2)证明:
方法一:∵y=-2x+b过C点,
∴b=6.
∴y=-2x+6.(4分)
∵当y=0时,x=3,
∴D(3,0).
∴BD=1.
∵OA=BC=2PO=BD=1,∠AOP=∠CBD,
∴△AOP≌△CBD.
∴∠PAO=∠DCB.
∵∠PAO+∠ACB=90°,
∴∠ACB+∠DCB=90°.
∴∠ACD=90°.
∴DC是⊙P的切线.(6分)
方法二:∵直线y=-2x+b过C点(2,2),
∴y=-2x+6.(4分)
又∵直线y=-2x+6交x轴于点D,y轴于点E,
∴D(3,0),E(0,6).
∴OD=3OE=6.
∴
.
又∵∠AOP=∠EOD,
∴△AOP∽△EOD.
∴∠APO=∠EDO.
又∵∠APO+∠PAO=90°,
∴∠EDO+∠PAO=90°.
∴∠ACD=90°.
∴CD是⊙O的切线.(6分)
(3)解:∵y=-
x2+mx+n过A(-2,0)和C(2,2),
∴![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231655335248962/SYS201310212316553352489003_DA/4.png)
解得
,
∴这个二次函数的解析式为y=-
x2+
x+3.(8分)
可求二次函数y=-
x2+
x+3与一次函数y=-2x+6的交点C(2,2)和D(3,0),
由此可知,满足条件的x的取值范围为2<x<3.(10分)
点评:此题考查的是用待定系数法求一次函数的解析式及圆的相关知识,涉及面较广.
(2)只要证明∠ACD=90°即可得到DC是⊙P的切线.
(3)把A,C两点代入解析式求出未知数的值,进而求出其解析式;可求二次函数y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231655335248962/SYS201310212316553352489003_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231655335248962/SYS201310212316553352489003_DA/1.png)
解答:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231655335248962/SYS201310212316553352489003_DA/images2.png)
∵OP⊥AB,
∴OB=OA=2.(1分)
∵OP2+AO2=AP2
∴OP2=5-4=1,OP=1,(2分)
∵AC是⊙P的直径,
∴∠ABC=90°.
∵CP=PA,BO=OA,
∴BC=2PO=2.
∴P(0,1),C(2,2).(3分)
(2)证明:
方法一:∵y=-2x+b过C点,
∴b=6.
∴y=-2x+6.(4分)
∵当y=0时,x=3,
∴D(3,0).
∴BD=1.
∵OA=BC=2PO=BD=1,∠AOP=∠CBD,
∴△AOP≌△CBD.
∴∠PAO=∠DCB.
∵∠PAO+∠ACB=90°,
∴∠ACB+∠DCB=90°.
∴∠ACD=90°.
∴DC是⊙P的切线.(6分)
方法二:∵直线y=-2x+b过C点(2,2),
∴y=-2x+6.(4分)
又∵直线y=-2x+6交x轴于点D,y轴于点E,
∴D(3,0),E(0,6).
∴OD=3OE=6.
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231655335248962/SYS201310212316553352489003_DA/2.png)
又∵∠AOP=∠EOD,
∴△AOP∽△EOD.
∴∠APO=∠EDO.
又∵∠APO+∠PAO=90°,
∴∠EDO+∠PAO=90°.
∴∠ACD=90°.
∴CD是⊙O的切线.(6分)
(3)解:∵y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231655335248962/SYS201310212316553352489003_DA/3.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231655335248962/SYS201310212316553352489003_DA/4.png)
解得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231655335248962/SYS201310212316553352489003_DA/5.png)
∴这个二次函数的解析式为y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231655335248962/SYS201310212316553352489003_DA/6.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231655335248962/SYS201310212316553352489003_DA/7.png)
可求二次函数y=-
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231655335248962/SYS201310212316553352489003_DA/8.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021231655335248962/SYS201310212316553352489003_DA/9.png)
由此可知,满足条件的x的取值范围为2<x<3.(10分)
点评:此题考查的是用待定系数法求一次函数的解析式及圆的相关知识,涉及面较广.
![](http://thumb2018.1010pic.com/images/loading.gif)
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