设直线方程是x=my+4圆心C1(-3,1)到直线的距离是d=|-3-m-4|/根号(1+m^2)又有勾股定理得到:d^2+(2根号3/2)^2=R^2即有(m+7)^2/(1+m^2)+3=4m^2+14m+49=m^2+114m=-48m=-24/7即直线方程是7x+24y=28