As the circle passes through points (9,8)(9,8) and (2,5)(2,5). the center lies on the perpendicular bisector joining these points as well as line y=1/6x+1y=16x+1
The slope of line joining (9,8)(9,8) and (2,5)(2,5) is (5-8)/(2-9)=3/75−82−9=37 and as such the slope of perpendicular bisector is -1/(3/7)=-7/3−137=−73.
As midpoint of line joining (9,8)(9,8) and (2,5)(2,5) is ((9+2)/2,(8+5)/2)(9+22,8+52) or (11/2,13/2)(112,132), the equation of perpendicular bisector is
y-13/2=-7/3(x-11/2)y−132=−73(x−112)
or 6y-39=-14x+776y−39=−14x+77
or 14x+6y=11614x+6y=116
Now putting value of yy from y=1/6x+1y=16x+1, we get
14x+6(1/6x+1)=11614x+6(16x+1)=116 or 15x=11015x=110 or x=22/3x=223
and hence y=22/3xx1/6+1=11/9+1=20/9y=223×16+1=119+1=209
Hence center of circle is (22/3,20/9)(223,209) and radius is distance between (22/3,20/9)(223,209) and (9,8)(9,8) i.e.
sqrt((9-22/3)^2+(8-20/9)^2)√(9−223)2+(8−209)2
= sqrt((5/3)^2+(52/9)^2)√(53)2+(529)2
= 1/9sqrt(225+2704)=sqrt2929/919√225+2704=√29299
and equation of circle is (x-22/3)^2+(y-20/9)^2=2929/81(x−223)2+(y−209)2=292981
or (9x-66)^2+(9y-20)^2=2929(9x−66)2+(9y−20)2=2929
or 81x^2+81y^2-1188x-360y+4356+400=292981x2+81y2−1188x−360y+4356+400=2929
or 81x^2+81y^2-1188x-360y+1827=081x2+81y2−1188x−360y+1827=0
or 9x^2+9y^2-132x-40y+203=09x2+9y2−132x−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]}
