Let the centre be (a,b), and the radius be ‘r’. the centre lies on x-y-1=0. therefore,

a = b+1.  equation of circle is (x-b-1)² +(y-b) ² = r²,  circle touches x-axis (y = 0).

Therefore, the quadratic equation in x: (x-b-1)² + b² = r², has only one solution. Therefore, its discriminant is zero. Solving the equation arising out of this condition, we get   b² = r².   (4x/3)-y+(4/3) =0, is also a tangent. From this we get

: (x-b-1)² + ((4x/3)+(4/3)-b) ² = r². from   b² = r², and setting the discriminant of this quadratic equation equal to zero, we get

3b²-2b-8 = 0. Thus b=2 or -4/3.  Centre lies in third quadrant. Therefore, b<0. Thus,

b=-4/3, a=b+1=-1/3, b² = r².=16/9. thus, equation of circle is

(x+(1/3))² + (y+(4/3)) ² = 16/9  

let the center of the circle be (h,k)   

   coordinates of center will satisfy x-y-1=0 

   so we get h-k-1=0......................eq 1

   a line is tangent to a circle if c=aroot(1+m^2) 

  here a is the radius of circle 

   putting values of c,m from the equation of tangent we get a=4/5

   perpendicular distance from the tangent to the center of the circle is (4h-3k+4)/5  which is equal to the radius of the circle so we get  4h-3k=0..........eq 2

  on solving eq 1 qnd 2 we get center as (-3,-4)

   eq of circle is (x+3)^2 +(y+4)^2=16/25

eq. is       (x+1/3)² + (y+4/3)²=(4/3)²