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

The general form of the equation of a circle is:

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

where #(x, y)# is any point on the circle, #(h, k)# is the center, and r is the radius.

Use equation [1] to write two equations, using the given points:

#(3 - h)^2 + (4 - k)^2 = r^2" [2]"#
#(6 - h)^2 + (1 - k)^2 = r^2" [3]"#

Temporarily eliminate r by setting the left side of equation [2] equal to the left side of equation [3]:

#(3 - h)^2 + (4 - k)^2 = (6 - h)^2 + (1 - k)^2#

Expand the squares, using the pattern #(a - b)^2 = a^2 - 2ab + b^2#:

#9 - 6h + h^2 + 16 - 8k + k^2 = 36 - 12h + h^2 + 1 - 2k + k^2#

The square terms cancel:

#9 - 6h + 16 - 8k = 36 - 12h + 1 - 2k#

Collect all of the constant terms into a single term on the right:

#-6h - 8k = -12h - 2k + 12#

Collect all of the h terms into a single term on the right:

# - 8k = -6h - 2k + 12#

Collect all of the k terms into a single term on the left:

# -6k = -6h + 12#

Divide both sides of the equation by -6:

#k = h - 2" [4]"#

Evaluate the given equation at the point #(h, k)#

#k = 2/3h + 7" [5]"#

Set the right side of equation [4] equal to the right side of equation [5]:

#h - 2 = 2/3h + 7#

#1/3h = 9#

#h = 27#

Substitute 27 for h in equation [5]:

#k = 2/3(27) + 7#

#k = 18 + 7#

#k = 25#

Substitute the center, #(27,25)#, into equation [2] and solve for r:

#(3 - 27)^2 + (4 - 25)^2 = r^2#

#(-24)^2 + (-21)^2 = r^2#

#r^2 = 1017#

#r = sqrt(1017) = 3sqrt(113)#

Substitute the center, #(27,25)#, and the radius, #3sqrt(113)# into equation [1]:

#(x - 27)^2 + (y - 25)^2 = (3sqrt(113))^2#

This is the equation of the circle.