A circle has a center that falls on the line #y = 2/9x +4 # and passes through # ( 3 ,1 )# and #(5 ,7 )#. What is the equation of the circle?

The equation for a circle:

#(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] and the two given points to write two equations:

#(3 - h)^2 + (1 - k)^2 = r^2" [2]"#
#(5 - h)^2 + (7 - k)^2 = r^2" [3]"#

Substitute h for x and k for y in the given equation:

#k = 2/9h + 4#

Multiply both sides of the equation by 9:

#9k = 2h + 36#

Write in standard form:

#2h - 9k = -36" [4]"#

Because #r = r# we can set the left side of equation [2] equal to the left side of equation [3]:

#(3 - h)^2 + (1 - k)^2 = (5 - h)^2 + (7 - k)^2#

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

#9 - 6h + h^2 + 1 - 2k + k^2 = 25 - 10h + h^2 + 49 - 14k + k^2#

The #h^2 and k^2# terms cancel:

#9 - 6h + 1 - 2k = 25 - 10h + 49 - 14k#

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

#-6h - 2k = -10h - 14k + 64#

Combine the h terms and the k terms on the left:

#4h + 12k = 64#

Divide both sides of the equation by 4:

#h + 3k = 16" [5]"#

Here are the two equations that define the values of h and k:

#2h - 9k = -36" [4]"#
#h + 3k = 16" [5]"#

Multiply equation [5] by 3 and add to equation [4]:

#5h = 12#

#h = 12/5#

Substitute #12/5# for h in equation [4]:

#2(12/5) - 9k = -36#

#9k = 36 + 24/5#

#k = 68/15#

To find the value of r, substitute the value for h and k into equation [3]:

#(5 - 12/5)^2 + (7 - 68/15)^2 = r^2"#

#(5 - 36/15)^2 + (7 - 68/15)^2 = r^2"#

#r = 17sqrt(10)/15#

Substitute these values into equation [1]:

#(x - 12/5)^2 + (y - 68/15)^2 = (17sqrt(10)/15)^2#