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

1 Answer
Narad T.
Oct 29, 2016

The equation of the circle is #(x-44/7)^2+(y-108/7)^2=208.3#

Explanation:

Let the center of the circle be #(a,b)# and radius r
Then the equation of the circle is
#(x-a)^2+(y-b)^2=r^2#

Then, #(1-a)^2+(2-b)^2=r^2#
and #(6-a)^2+(1-b)^2=r^2#

Equating the equations
#(1-a)^2+(2-b)^2=(6-a)^2+(1-b)^2#
developing,
#1-2a+a^2+4-4b+b^2=36-12a+a^2+1-2b+b^2#

#5-2a-4b=37-12a-2b#
#10a-2b=32#
#5a-b=16# this is equation 1
Also, #y=(3x)/2+6#
So #b=(3a)/2+6#
Solving for a and be, we get #a=44/7# and #b=108/7#
We calculate the radius #r^2=(6-44/7)^2+(1-108/7)^2#

#r^2=(2/7)^2+(108/7)^2##=># #r=14.43#

Equation of circle is
#(x-44/7)^2+(y-108/7)^2=208.3#

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