(1)方法一:由题意,原点O必定在圆M内,即点(0,0)代入方程x
2+y
2+Dx+Ey+F=0的左边所得的值小于0,于是有F<0,即证.
方法二:由题意,不难发现A,C两点分别在x轴正、负半轴上.设两点坐标分别为A(a,0),C(c,0),则有ac<0.对于圆的方程x
2+y
2+Dx+Ey+F=0,当y=0时,可得x
2+Dx+F=0,其中方程的两根分别为点A和点C的横坐标,于是有x
Ax
C=ac=F.
因为ac<0,故F<0.
(2)不难发现,对角线互相垂直的四边形ABCD的面积S=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407032457.png)
,因为S=8,|AC|=2,可得|BD|=8.
又因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407016371.png)
·
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407032376.png)
=0,所以∠BAD为直角,又因为四边形是圆M的内接四边形,故|BD|=2r=8⇒r=4.
对于方程x
2+y
2+Dx+Ey+F=0所表示的圆,
可知
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407094371.png)
+
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407110374.png)
-F=r
2,所以D
2+E
2-4F=4r
2=64.
(3)设四边形四个顶点的坐标分别为A(a,0),B(0,b),C(c,0),D(0,d).
则可得点G的坐标为(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407126276.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407141308.png)
),即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407157379.png)
=(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407126276.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407141308.png)
).
又
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407016371.png)
=(-a,b),且AB⊥OH,故要使G,O,H三点共线,只需证
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407016371.png)
·
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407157379.png)
=0即可.
而
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407016371.png)
·
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407157379.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407282404.png)
,且对于圆M的一般方程x
2+y
2+Dx+Ey+F=0,
当y=0时可得x
2+Dx+F=0,其中方程的两根分别为点A和点C的横坐标,
于是有x
Ax
C=ac=F.
同理,当x=0时,可得y
2+Ey+F=0,其中方程的两根分别为点B和点D的纵坐标,于是有y
By
D=bd=F.
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407016371.png)
·
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407157379.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824040407282404.png)
=0,即AB⊥OG.
故O,G,H三点必定共线.