A straight line l2 with a center is perpendicular to L, and the slope of l2 is the negative reciprocal of L, which is-1.

So the l2 equation is: x+y-7=0, and the intersection of L and l2 is point P.

P (4,3) can be obtained by simultaneous equations.

Because l 1 intersects with point p, the equation of l 1 can be set as y-3=k(x-4), that is, kx-y+3-4k=0(A=K, B=- 1, C=3-4K).

Because PQ=2 and r= root number 2, the distance from the center of the circle to l 1 is two thirds of d= the square of root number 2 minus root number = 1.

The absolute value of d = 1=(AX+BY+C) divided by the root sign (Party A+Party B).

Substitute A=K, B=- 1, C=3-4K and x=4, y=3 to get k=0.

l 1:y-3=0