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

The equation of a circle is:

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

where #(h,k)# is the center and r is the radius.

Substitute the point (h,k) into the equation for the line on which the center lies:

#k = 3/2h + 1##" [1]"#

We can use the two given points and the equation of a circle to write two more equations:

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

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

We can, temporarily, eliminate the variable r by setting the left side of equation [2] equal to the left side of equation [3]:

#(1 - h)^2 + (2 - k)^2 = (6 - h)^2 + (1 - k)^2#

Use the pattern #(a - b)^2 = a^2 - 2ab + b^2#, to expand the squares:

#1 - 2h + h^2 + 4 - 4k + k^2 = 36 - 12h + h^2 + 1 - 2k + k^2#

The square terms are the same on both sides of the equation, therefore, they cancel:

#1 - 2h + 4 - 4k = 36 - 12h + 1 - 2k#

Collect the 4 constant terms into a single term on the right:

#-2h - 4k = 32 - 12h - 2k#

Collect the 2 h terms into a single term on the right:

#-4k = 32 - 10h - 2k#

Collect the 2 k terms into a single term on the left:

#-2k = 32 - 10h#

Divide both sides by -2:

#k = 5h - 16##" [4]"#

Subtract equation [4] from equation [1]:

#k - k = 3/2h - 5h + 1 + 16#

#0 = -7/2h + 17#

#h = 34/7#

Substitute 34/7 for h in equation [1]:

#k = 3/2(34/7) + 1#

#k = 58/7#

Substitute the values for h and k into equation [2] or [3] (I will use equation [2]):

#(1 - 34/7)^2 + (2 - 58/7)^2 = r^2#

#(-27/7)^2 + (-44/7)^2 = r^2#

#r^2 = 2665/7^2#

#r = sqrt(2665)/7#

The equation of the circle is:
#(x - 34/7)^2 + (y - 58/7)^2 = (sqrt2665/7)^2#