Let #(x_1, 0)# be the center on the x-axis & #r# be the radius of circle then the equation of circle
#(x-x_1)^2+(y-0)^2=r^2#
#(x-x_1)^2+y^2=r^2#
Now, since the above circle is tangent to the line #x+y=2# at #(4, -2)# hence the point #(4, -2)# will satisfy the equation of circle
#(4-x_1)^2+(-2)^2=r^2#
#r^2=x_1^{2}-8x_1+20\ .....(1)#
Now, the line #x+y=2# is tangent to the circle hence, substituting #y=2-x# in equation of circle we get
#(x-x_1)^2+(2-x)^2=r^2#
#2x^2-2(x_1+2)x+x_1^{2}+4-r^2=0#
The above equation will have equal roots iff discriminant #B^2-4AC=0#
#(-2(x_1+2))^2-4(2)(x_1^{2}+4-r^2)=0#
#x_1^{2}-4x_1+4-2r^2=0#
substituting the value of #r^2# from (1) in above equation, we get
#x_1^{2}-4x_1+4-2(x_1^{2}-8x_1+20)=0#
#x_1^{2}-12x_1+36=0#
#(x_1-6)^2=0#
#x_1=6, 6#
we get sinhle real value of #x# i.e. #x=6#, substituting this value in (1), we get
#r^2=6^{2}-8\cdot 6+20#
#r^2=8#
hence, the equation od circle is
#(x-6)^2+y^2=8#
#x^2+y^2-12x+28=0#
