A circle has a center that falls on the line #y = 8/9x +4 # and passes through # ( 3 ,1 )# and #(5 ,7 )#. What is the equation of the circle?

1 Answer
Oct 26, 2016

The equation of the circle is
#(x-1.091)^2+(y-4.97)^2=4.41^2#

Explanation:

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

The circle passes through the two points. So
#(3-a)^2+(1-b)^2=r^2#
and
#(5-a)^2+(7-b)^2=r^2#

We can equalize the two equations

#(3-a)^2+(1-b)^2=(5-a)^2+(7-b)^2#
Developing and simplifying
#9-6a+a^2+1-2b+b^2=25-10a+a^2+49-14b+b^2#

#10-6a-2b=74-10a-14b#

#12b=-4a+64# #=># #3b=-a+16#, this equation 1

We obtain the equation 2 from the equation of the line
#b=(8a)/9+4#

Solving for #a# and #b#, we get #a=1.091# and #b=4.97#
and the radius of the circle is

#r^2=(3-1.091)^2+(1-4.97)^2#
#r^2=1.909^2+3.97^2#

#r=4.41#
So the equation of the circle is
#(x-1.091)^2+(y-4.97)^2=4.41^2#