圆心在直线3x-y=0上,y=3x,设圆心坐标为（m,3m),它与X轴相切,则半径为3m,设圆方程为：(x-m)^2+(y-3m)^2=9m^2,设直线y=x与圆相交于A(x1,y1),B(x2,y2),x1=y1,x2=y2,y=x代入圆方程,2x^2-8mx+m^2=0,根据韦达定理,x1+x2=4m...