According to the tangent of the straight line and the circle, find the straight line with the center of the circle:

sqrt(3)* x-y-4 * sqrt(3)= 0( 1)；

Let the center of the circle be (x 1, y 1), according to the above formula:

sqrt(3)* x 1-y 1-4 * sqrt(3)= 0(2)；

Furthermore, the distance from the center of the circle (x 1, y 1) to the tangent point is equal to subtracting 1 from the center of the circle (1, 0) to get the second equation:

sqrt[(x 1- 1)^2+y 1^2)- 1 = sqrt[(x 1-3)^2+(y+sqrt(3))^2](3)

The above three equations are solved simultaneously: x 1=4, y1= 0; r = 4-2 = 2；

The equation of a circle is: (x-4) 2+y 2 = 4.