1.解答;设A(x1;y1)B(x2;y2)他们的中点为M（x;y)则x=(x1+x2)/2 ,y=(y1+y2)向量pA=(x1-4,y1-4)向量PB=（x2-4,y2-4)
则有（x1-4)(x2-4)+(y1-4)(y2-4)=0即x1x2+y1y2-4（x1+x2)-4(y1+y2)+32=0因为A,B都在圆上。故有x1^2+y1^2=36x2^2+y2^2=36两式相加得x1^2+x2^2+y1^2+y2^2=72即（x1+x2)^2+(y1+y2)^2-72=2(x1x2+y1y2)由以上式子消去x1,x2,y1,y2得M的轨迹为（x-2)^2+(y-2)^2=10 |
