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

From the given two points #(3, 7)# and #(7, 1)# we will be able to establish equations

#(x-h)^2+(y-k)^2=r^2#

#(3-h)^2+(7-k)^2=r^2" "#first equation using #(3, 7)#

and

#(x-h)^2+(y-k)^2=r^2#

#(7-h)^2+(1-k)^2=r^2" "#second equation using #(7, 1)#

But #r^2=r^2#
therefore we can equate first and second equations

#(3-h)^2+(7-k)^2=(7-h)^2+(1-k)^2#

and this will be simplified to
#h-3k=-2" "#third equation
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The center #(h, k)# passes thru the line #y=1/3x+7# so we can have an equation

#k=1/3h+7# because the center is one of its points

Using this equation and the third equation,

#h-3k=-2" "#
#k=1/3h+7#

The center #(h, k)=(19, 40/3)# by simultaneous solution.

We can use the equation
#(3-h)^2+(7-k)^2=r^2" "#first equation
to solve for the radius #r#

#r^2=2665/9#

and the equation of the circle is

#(x-19)^2+(y-40/3)^2=2665/9#

Kindly see the graph to verify the equation of the circle #(x-19)^2+(y-40/3)^2=2665/9# colored red, with points #(3, 7)# colored green, and #(7, 1)# colored blue, and the line #y=1/3x+7# colored orange which contains the center #(19, 40/3)# colored black.

God bless....I hope the explanation is useful.