A circle has a center that falls on the line #y = 2/7x +7 # and passes through # ( 3 ,4 )# and #(6 ,1 )#. What is the equation of the circle?

The general Cartesian form for the equation of a circle is:

#(x-h)^2 + (y-k)^2 = r^2" [1]"#

Substitute the point #(3,4)# into equation [1]:

#(3-h)^2 + (4-k)^2 = r^2" [2]"#

Substitute the point #(6,1)# into equation [1]:

#(6-h)^2 + (1-k)^2 = r^2" [3]"#

Expand the squares of equation [2] and [3]:

#9-6h+h^2 + 16-8k+k^2 = r^2" [4]"#
#36-12h+h^2 + 1-2k+k^2 = r^2" [5]"#

Subtract equation [5] from equation [4]:

#color(white)(....)9-6h+h^2 + 16-8k+k^2 = r^2" [4]"#
#ul(-36+12h-h^2 - 1+2k-k^2 = -r^2)" [5]"#
#-27+6h+0h^2+15-6k+0k^2= 0r^2#

Remove all of the 0 terms and move the constant to the right:

#6h - 6k = 12" [6]"#

Evaluate the given equation, #y = 2/7x +7#, at the center #(h,k)#:

#k = 2/7h+7" [7]"#

Substitute equation [7] into equation [6]:

#6h - 6(2/7h+7) = 12#

Solve for h:

#42h -12h- 294 = 84#

#30h = 378#

#h = 63/5#

Use equation [7] to find the value of k:

#k = 2/7(63/5)+7#

#k = 53/5#

Use equation [2] to find the value of r:

#(3-63/5)^2 + (4-53/5)^2 = r^2#

#(15/5-63/5)^2 + (20/5-53/5)^2 = r^2#

#(-48/5)^2 + (-33/5)^2 = 3393/25 = r^2#

#r = (3sqrt377)/5#

Use equation [1] to write the equation:

#(x - 63/5)^2 + (y - 53/5)^2 = ((3sqrt377)/5)^2#