A circle has a center that falls on the line #y = 12/7x +3 # and passes through # ( 9 ,5 )# and #(8 ,7 )#. What is the equation of the circle?

1 Answer
Narad T.
Oct 23, 2016

The center of the circle is #(-7/11,91/44)#
And the radius is #9.95#
The equation of the circle is #(x+7/11)^2+(y-91/44)^2=98.9#

Explanation:

Let #(a,b)# be the center of the circle and #r# the radius
The equation of a circle is #(x-a)^2+(y-b)^2=r^2#
So we apply this to the 2 points and we get

#(9-a)^2+(5-b)^2=r^2#
#(8-a)^2+(7-b)^2=r^2#
So we can combine these 2 equations
#(9-a)^2+(5-b)^2=(8-a)^2+(7-b)^2#
Developing

#81-18a+a^2+25-10b+b^2=64-16a+a^2+49-14b+b^2#

Simplifying
#81-18a+25-10b=64+49-16a-14b#
#106-113=18a-16a-14b+10b#
#-7=-2a-4b# #=># #4b=-2a+7# this equation #1#
From the equation of the line, we get
#y=(12x)/7+3# #=># #b=(12a)/7+3# this is equation #2#

So#-a/4+7/4=(12a)/7+3##=># #a=-7/11#
and #4b=-2*-7/11+7##=># #b=91/44#

from these we can calculate #r#
#r=sqrt((8--7/11)^2+(7-91/44)^2)=9.95#

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