As a perpendicular drawn from center of a circle bisects the chord, the center of circle also falls on the perpendicular bisector of lie joining #(3,1)# and #(5,7)#. Hence we should forst find the equation of their perpendicular bisector.
Mid point of #(3,1)# and #(5,7)# is #((3+5)/2,(1+7)/2)# i.e. #(4,4)#. And as slope of line joining #(3,1)# and #(5,7)# is #(7-1)/(5-3)=3# and that of perpendicular bisector would be #-1/3#. As it passes through #(4,4)#, its equation is
#y-4=-1/3(x-4)# or #3y-12=-x+4# i.e. #x+3y=16#
and center of circle falls on intersection of #x+3y=16# and line #y=5/9x+4#. Their solution should give us the center and hence putting #y=5/9x+4# in #x+3y=16#, we get
#x+3xx(5/9x+4)=16# or #x+5/3x+12=16#
i.e. #8/3x=4# and #x=4xx3/8=3/2# and hence
#y=5/9xx3/2+4=5/6+4=29/6# i.e. center of circle is #(3/2,29/6)#
and as it passes through #(3,1)# redius of circle is
#sqrt((3-3/2)^2+(29/6-1)^2)=sqrt(9/4+529/36)=sqrt610/36)#
and equation of circle is #(x-3/2)^2+(y-29/6)^2=610/36#
or #9(2x-3)^2+(6y-29)^2=610#
or #36x^2+36y^2-108x-348y+81+841=610#
or #36x^2+36y^2-108x-348y+312=0#
i.e. #3x^2+3y^2-9x-29y+26=0#
graph{(x+3y-16)(9y-5x-4)((x-3)^2+(y-1)^2-0.03)((x-5)^2+(y-7)^2-0.03)(3x^2+3y^2-9x-29y+26)=0 [-8.875, 11.125, -0.4, 9.6]}
