试题分析:(1)连接CB,CO,则CB∥y轴,由圆周角定理、勾股定理得OC=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823033337895661.png)
,则半径OO′=5,S⊙O′=π•5
2=25π.
(2)过点A作AD⊥x轴于点D,依题意,得∠BAD=30°,在Rt△ABD中,设BD=x,则AB=2x,由勾股定理AD=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823033337910344.png)
x,根据图形得到OD=OB+BD=6+x,故AB=2x=6(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823033337910344.png)
+1)≈16.2
(3)过点A作AG⊥y轴于点G.过点O′作O′E⊥OB于点E,并延长EO′交AG于点F.由垂径定理得,OE=BE=3.在Rt△OO′E中,由勾股定理得,O′E=4.所以O′F=9+3
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823033337910344.png)
-4=5+3
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823033337910344.png)
>5.
(1)连接CB,CO,则CB∥y轴,
∴∠CBO=90°,
设O′为由O、B、C三点所确定圆的圆心.
则OC为⊙O′的直径.
由已知得OB=6,CB=8,由勾股定理得OC=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823033337895661.png)
半径OO′=5,S⊙O′=π•5
2=25π.
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082303333798811819.png)
(2)过点A作AD⊥x轴于点D,依题意,得∠BAD=30°,
在Rt△ABD中,设BD=x,则AB=2x,
由勾股定理得,AD=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823033338004826.png)
,
由题意知:OD=OB+BD=6+x,在Rt△AOD中,OD=AD,6+x=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823033337910344.png)
x
∴x=3(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823033337910344.png)
+1),
∴AB=2x=6(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823033337910344.png)
+1)≈16.2
(3)过点A作AG⊥y轴于点G.
过点O′作O′E⊥OB于点E,并延长EO′交AG于点F.
由(1)知,OO′=5,由垂径定理得,OE=BE=3.
∴在Rt△OO′E中,由勾股定理得,O′E=4
∵四边形FEDA为矩形.
∴EF=DA,而AD=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823033337910344.png)
x=9+3
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823033337910344.png)
∴O′F=9+3
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823033337910344.png)
-4=5+3
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823033337910344.png)
>5,
∴直线AG与⊙O′相离,A船不会进入海洋生物保护区.
考点: 1.勾股定理的应用;2.点与圆的位置关系.