As the circle passes through points #(9,8)# and #(2,5)#. the center lies on the perpendicular bisector joining these points as well as line #y=1/6x+1#
The slope of line joining #(9,8)# and #(2,5)# is #(5-8)/(2-9)=3/7# and as such the slope of perpendicular bisector is #-1/(3/7)=-7/3#.
As midpoint of line joining #(9,8)# and #(2,5)# is #((9+2)/2,(8+5)/2)# or #(11/2,13/2)#, the equation of perpendicular bisector is
#y-13/2=-7/3(x-11/2)#
or #6y-39=-14x+77#
or #14x+6y=116#
Now putting value of #y# from #y=1/6x+1#, we get
#14x+6(1/6x+1)=116# or #15x=110# or #x=22/3#
and hence #y=22/3xx1/6+1=11/9+1=20/9#
Hence center of circle is #(22/3,20/9)# and radius is distance between #(22/3,20/9)# and #(9,8)# i.e.
#sqrt((9-22/3)^2+(8-20/9)^2)#
= #sqrt((5/3)^2+(52/9)^2)#
= #1/9sqrt(225+2704)=sqrt2929/9#
and equation of circle is #(x-22/3)^2+(y-20/9)^2=2929/81#
or #(9x-66)^2+(9y-20)^2=2929#
or #81x^2+81y^2-1188x-360y+4356+400=2929#
or #81x^2+81y^2-1188x-360y+1827=0#
or #9x^2+9y^2-132x-40y+203=0#
graph{(9x^2+9y^2-132x-40y+203)((x-9)^2+(y-8)^2-0.03)((x-2)^2+(y-5)^2-0.03)(6y-x-6)(14x+6y-116)=0 [-3.95, 16.05, -1.28, 8.72]}
