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]}
