∴16+4+4D+2E+F=0,0②

　　1+9-D+3E+F=0.0③

　　由①②③解得D=-2,E=0,F=-12.

　　故所求圆的方程为x2+y2-2x-12=0.

12.导学号57084087圆C过点A(6,0),B(1,5),且圆心在直线l:2x-7y+8=0上.

(1)求圆C的方程;

(2)P为圆C上的任意一点,定点Q(8,0),求线段PQ中点M的轨迹方程.

解(1)(解法一)直线AB的斜率k=(5"-" 0)/(1"-" 6)=-1,

　　所以线段AB的垂直平分线m的斜率为1.

　　线段AB的中点的横坐标和纵坐标分别为x=(6+1)/2=7/2,y=(0+5)/2=5/2.因此,直线m的方程为y-5/2=x-7/2,即x-y-1=0.

　　又圆心在直线l上,所以圆心是直线m与直线l的交点.联立方程组{■(x"-" y"-" 1=0"," @2x"-" 7y+8=0"," )┤解得{■(x=3"," @y=2"," )┤

　　所以圆心坐标为C(3,2).又半径r=|CA|=√13,

　　则所求圆的方程是(x-3)2+(y-2)2=13.

　　(解法二)设所求圆的方程为(x-a)2+(y-b)2=r2.

　　由题意得{■("(" 6"-" a")" ^2+"(" 0"-" b")" ^2=r^2 "," @"(" 1"-" a")" ^2+"(" 5"-" b")" ^2=r^2 "," @2a"-" 7b+8=0"," )┤解得{■(a=3"," @b=2"," @r^2=13"," )┤

　　所以所求圆的方程是(x-3)2+(y-2)2=13.

　　(2)设线段PQ的中点M(x,y),P(x0,y0),

　　则{■((x_0+8)/2=x"," @(y_0+0)/2=y"," )┤解得{■(x_0=2x"-" 8"," @y_0=2y"," )┤将P(2x-8,2y)代入圆C中,得(2x-8-3)2+(2y-2)2=13,即线段PQ中点M的轨迹方程为(x"-" 11/2)^2+(y-1)2=13/4.

13.导学号57084088设△ABC的顶点坐标A(0,a),B(-√3a,0),C(√3a,0),其中a>0,圆M为△ABC的外接圆.

(1)求圆M的方程.

(2)当a变化时,圆M是否过某一定点?请说明理由.

解(1)设圆M的方程为x2+y2+Dx+Ey+F=0.

　　∵圆M过点A(0,a),B(-√3a,0),C(√3a,0),

　　∴{■(a^2+aE+F=0"," @3a"-" √3a D+F=0"," @3a+√3a D+F=0"," )┤

解得D=0,E=3-a,F=-3a.