A circle has a center that falls on the line #y = 8/9x +4 # and passes through # ( 4 ,1 )# and #(3 ,7 )#. What is the equation of the circle?

So the center C of the circle falls on the line #y=(8x+36)/9#
If we find the equation of other line in which a radius of the circle stands, this two lines will meet at the circle's center because the circle's radii converge to its center.

Calling
#A(4,1)#
#B(3,7)#
The segment AB is a chord of the circle that is divided in the precise middle by a radius that is perpendicular to this chord.

#M(3.5, 4)#
Slope of the line in which segment AB stands and of a line perpendicular to it
#k=(Delta y)/(Delta x)=(7-1)/(3-4)=6/(-1)=-6# => #p=-1/k=1/6#

Other equation in which a radius stands:
#MC -> y-4=(1/6)(x-3.5)# => #y=(x-3.5)/6+4# => #y=(x+20.5)/6#

Finding the intersection of the two lines:
#(8x+36)/9=(x+20.5)/6# => #48x+216=9x+184.5# => #39x=-31.5# => #x=-63/78#
#-> y=(-63/78+41/2)/6=(-63+1599)/468=1536/468# => #y=128/39#
=> #C(-63/78,128/39)#

Finding the radius
#r=sqrt((4+63/78)^2+(1-128/39)^2)=(sqrt((312+63)^2+4*(39-128)^2))/78=sqrt(140625+31684)/78=sqrt(172309)/78~=5.414#

Then the equation of the circle is
#(x-x_c)^2+(y-y_c)=r^2#
#(x+63/78)^2+(y-128/39)^2=172309/6084#