As the circle passes through #(2,8)# and #(3,9)#, its centre must lie on perpendicular bisector of the line joining these two.
Hence, the centre lies on a line passing through their midpoint i.e. #((2+3)/2,(8+9)/2)# i.e. #(5/2,17/2)#,
and slope of line joining #(2,8)# and #(3,9)# is #(9-8)/(3-2)=1#, slope of perpendicular bisector would be #(-1)/1=-1# and its equation is
#(y-17/2)=-1xx(x-5/2)# i.e. #x+y=11#
Hence, centre lies on point of intersection of #x+y=11# and #y=8/7x+2#. Putting the latter value of #y# in former, we get
#x+8/7x+2=11# i.e. #15/7x=9# and #x=9xx7/15=21/5#
and #y=11-21/5=34/5# and centre of circle is #(21/5,34/5)#. The radius of circle would be distance of this centre from a point on circle say #(2,8)# i.e.
#sqrt((8-34/5)^2+(2-21/5)^2)=sqrt(36/25+121/25)=sqrt(157/25)#
and equation of circle is
#(x-21/5)^2+(y-34/5)^2=157/25#
or #x^2-42/5x+441/25+y^2-68/5y+1156/25=157/25#
or #25x^2+25y^2-210x-340y+1440=0#
or #5x^2+5y^2-42x-68y+288=0#
graph{(5x^2+5y^2-42x-68y+288)(x+y-11)(8x-7y+14)((x-2)^2+(y-8)^2-0.01)((x-3)^2+(y-9)^2-0.01)=0 [-1.873, 8.127, 4.46, 9.46]}
