Let the equation of the circle be
#x^2+y^2+2gx+2fy+c = 0#
It is a standard result that the tangent of this circle at the point #(x_t,y_t)# is given by
# x x_t +y y_t +g(x+x_t)+f(y+y_t)+c = 0 #
which can be written in the form
#(x_t+g)x+(y_t +f)y+gx_t+fy_t +c = 0#
Now we know that #y=1# is tangent to this circle at some point #(x_1, y_1)# . Comparison allows us to write
#x_1+g = 0, y_1+f = 1, gx_1 +fy_1 +c =0#
Eliminating #x_1# and #y_1# gives the equation
#-g^2+f(1-f)+c = 0 #
Again, we know that #y=x# is tangent to this circle at some point #(x_2, y_2)#. This gives
#x_2+g = 1, y_2+f = -1, gx_2 +fy_2 +c =0#
Eliminating #x_2# and #y_2# :
# g(1-g)-f(1+f) +c = 0 #
Finally, tangency to #x+y-6 = 0# gives
#x_3+g = 1, y_3+f = 1, gx_3 +fy_3 +c =-6#
so that,
# g(1-g) +f(1-f)+c = -6#
So, we finally arrive at three equations
(a) #-g^2+f(1-f)+c = 0 #
(b) # g(1-g)-f(1+f) +c = 0 #
(c) # g(1-g) +f(1-f)+c = -6#
which we must solve for the three parameters #f#, #g# and #c#
Subtracting (a) from (b) gives
(d) #g=2f#,
and substituting this back in (a) leads to
#-(2f)^2+f(1-f)+c = 0 implies#
(e) #c = 5f^2-f#
Substituting both (d) and (e) in (c) gives, after some cancellations,
#f=3#
Thus #g =6# and #c= 5\times 3^2-3 = 42#
