(x- 1)2+y ^ 2 = 16。 Its center is (1, 0) and its radius is 4.

Since x=-3 must be one of the tangents, a circle can be made with A as the center and AB(y=3) as the radius, and the intersection of the two circles is the tangent point.

The equation of circle a is:

(x+3)^2+(y-3)^2=9.

Solve the equation group composed of two circular equations (two kinds of subtraction), and get:

8x-6y+24=0。 namely

4x-3y+ 12=0

Or y=4/3x+4.

This equation is an equation connecting two tangent points. By substituting this linear equation into the equation of any two circles, the tangent coordinates can be obtained.

The line segment where l intersects the circle is the largest, that is, l passes through the center of the circle. Let the l equation be y = kx+b.

Substituting point (-3,3) (1,0) gives:

k=-3/4，b=3/4

The l equation is y=-3/4x+3/4.