A circle has a center that falls on the line #y = 5/2x +1 # and passes through #(8 ,2 )# and #(3 ,1 )#. What is the equation of the circle?

1 Answer
Narad T.
Oct 21, 2016

The equatio of the circle is #(x-56/15)^2+(y-31/3)^2=87.65#

Explanation:

Let the center of the cicle be #(a,b)#
As the line passes through the center #b=(5a)/2+1#
The equation of the circle is #(x-a)^2+(y-b)^2=r^2#
where #r# is the radius
As the circle passes through #(8,2)# and (3,1)
we get, #(8-a)^2+(2-b)^2=r^2#
and #(3-a)^2+(1-b)^2=r^2#
Equating #(8-a)^2+(2-b)^2=(3-a)^2+(1-b)^2#
Developing
#64-16a+a^2+4-4b+b^2=9-6a+a^2+1-2b+b^2#

#64-16a+4-4b=9-6a-2b+1#

#68-16a-4b=10-6a-2b#

#58=10a+2b#

#5a+b=29# we compare this to the first equation
#b=(5a)/2+1#
#29-5a=(5a)/2+1#
#15a/2=28#
#a=2*28/15=56/15#
and #b=5/2*56/15+1=31/3#
So the radius is #r^2=(3-56/15)^2+(1-31/3)^2=11^2/15^2+28^2/9=87.65#
and #r=9.36#

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