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]}
