Suppose the garden: (x-a) 2+(y-b) 2 = r 2, the center of the circle (a, b), the radius r,

According to the known conditions, there are:

1.P( 1, 1) is on the straight line L 1: 4x-3y- 1 = 0, and the straight line is tangent to the garden, so there must be p as the tangent point.

2. The slope of the straight line l 1 and l2 is equal to = > L1/L2. When two parallel lines are tangent to a circle, there must be: the distance between the two straight lines is equal to the diameter 2r.

Distance from P( 1, 1) to line l2: 4x-3y+4=0:

2r = | 4-3+4 |/√( 16+9)= 1，r= 1/2，

From the center of the circle (a, b) to P( 1, 1), the distance of the straight line l2 is equal to =r,

(a- 1)^2+(b- 1)^2=( 1/2)^2- 1

| 4a-3 b+4 |/√( 16+9)= 1/2-2

The solution of 1 2 equation;

a=3/5，b= 13/ 10，

The equation of a circle is: (x-3/5) 2+(y-13/10) 2 =1/4.

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