As circle passes through #(3,5)# and #(1,4)#, it is equidistant from these two points and hence locus of such points is given by
#(x-3)^2+(y-5)^2=(x-1)^2+(y-4)^2# or
#x^2-6x+9+y^2-10y+25=x^2-2x+1+y^2-8y+16# or
#-6x+2x-10y+8y+9+25-1-16=0# or
#-4x-2y+17=0# or #4x+2y=17#
As it lies on #y=2/9x+8# and solving these two simultaneous equations should give us coordinates of the center. So putting this value of #y# in first equatin, we get
#4x+2(2/9x+8)=17# or #36x+2(2x+72)=153# (multiplying by #9#)
#36x+4x+144=153# or #40x=9# or #x=9/40#
and #y=2/9xx9/40+8=1/20+8=161/20#
Hence center is #(9/40,161/20)#
and the equation of circle is
#(x-9/40)^2+(y-161/20)^2=(9/40-3)^2+(161/20-5)^2# or
#x^2-9/20x+(9/40)^2+y^2-161/10y+(161/20)^2=(9/40)^2-27/20+9+(161/20)^2+25-805/10# or
#x^2-9/20x+y^2-161/10y=-27/20+9+25-805/10# or
#x^2-9/20x+y^2-161/10y=34-1647/20# or
#20x^2+20y^2-9x-322y=680-1647=-967# or
#20x^2+20y^2-9x-322y+967=0#
graph{(20x^2+20y^2-9x-322y+967)(y-2/9x-8)=0 [-9.75, 10.25, 2.8, 12.8]}
