设直线方程为y=kx+3k+4,即kx-y+3k+4=0
且与圆x^2+y^2=25相切
所以圆心（0,0）到直线的距离为半径5
|3k+4|/[根号（k^2+1）]=5
解得k=3/4
y=3/4*x+25/4
即3x-4y+25=0