设圆心为（x,y)
x^2+y^2-6mx-2(m-1)y+10m^2-2m-24=0
x^2-6mx+9m^2+y^2-2(m-1)+(m-1)^2=25
(x-3m)^2+(y-m+1)^2=25
x=3m ,y=m-1
m=x/3 ,m=y+1
所以x/3=y+1即x-3y-3=0