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

1 Answer
Narad T.
Oct 24, 2016

The equation of the circle is #(x-27/16)^2+(y-101/16)^2=39.2#

Explanation:

Let the center of the circle be #(a,b)#
Then the equation of the circle is #(x-a)^2+(y-b)^2=r^2#
As the circles passes through #(7,3)# and #(5,1)#
Then substituting those points in the equation of the circle

#(7-a)^2+(3-b)^2=r^2#
#(5-a)^2+(1-b)^2=r^2#

so #(7-a)^2+(3-b)^2=(5-a)^2+(1-b)^2#
Developing both sides
#49-14a+a^2+9-6b+b^2=25-10a+a^2+1-2b+b^2#
simplifying
#58-14a-6b=26-10a-2b#
#4a+4b=32#
#a+b=8# this is equation 1
Putting #(a,b)# in the equation of the line
#b=7/9a+5# this is equation 2
Solving for a and b
we obtain #(27/16,101/16)# as the center of the circle
The radius #r=sqrt (85^2+63^2)/16=6.26#
The equation of the circle is #(x-27/16)^2+(y-101/16)^2=39.2#

Related questions
See all questions in Equation of a Circle
Impact of this question
1223 views around the world
You can reuse this answer
Creative Commons License