Let #C(h,k)# be the centre, and, #r# the radius of the reqd. circle #S#.

The given pt.#P(1,1) in S rArr CP^2=r^2 rArr (h-1)^2+(k-1)^2=r^2...(1)#

#C in [x+y=3] rArr h+k=3..........(2)#

Finally, #S# touches #"X-axis : y=0"#

#":. the bot#-distance from #C(h,k)# to the #"X-axis"=r#.

#:. |k|=r...............(3)#

From #"(1) & (3)", (h-1)^2+(k-1)^2=k^2#.

BY #(2)#, then, #(3-k-1)^2+(k-1)^2-k^2=0#, or,

#(2-k)^2+(k-1)^2-k^2=0#

#k^2-6k+5=0 rArr (k-5)(k-1)=0 rArr k=5,or, 1#.

Accordingly, #"h=-2, or, 2, and, finally, r=5, or, 1"#.

Thus, we have #2# circles #S_1, and, S_2#:

# S_1 : Centre C_1(-2,5), r=5#, with the eqn. given by,

# S_1 : (x+2)^2+(y-5)^2=5^2, i.e., x^2+y^2+4x-10y+4=0#,

and,

#S_2 : Centre C_2(2,1), r=1, eqn. x^2+y^2-4x-2y+4+0"#.

Enjoy Maths.!