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

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Answer 1 · 2016-10-19T21:59:13

#(x - 38/3)^2 + (y - 142/3)^2 = (sqrt(19721)/3)^2#

The standard form for the equation of a circle is:

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

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

Using the two points, we write two equations:

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

Using the pattern #(a - b)^2 = a^2 - 2ab + b^2# we expand the squares:

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

Subtract the first equation from the second:

#35 -10h - 3 + 2k = 0#

#32 -10h + 2k = 0#

#k = 5h - 16#

Substitute #(h, k)# into the given equation:

#k = 7/2h + 3#

Set the right sides equal:

#5h - 16 = 7/2h + 3#

#10h - 32 = 7h + 6#

#h = 38/3#
#k = 142/3#

Substitute the center into the first equation:

#(1 - 38/3)^2 + (2 - 142/3)^2 = r^2#

#r = sqrt(19721)/3#