A circle has a center that falls on the line #y = 2/7x +3 # and passes through # ( 2 ,8 )# and #(3 ,5 )#. What is the equation of the circle?

1 Answer
Jun 30, 2017

The equation of the circle is #(x+56)^2+(y+13)^2=3805#

Explanation:

Let #C# be the mid point of #A=(2,8)# and #B=(3,5)#

#C=((2+3)/2,(8+5)/2)=(5/2,13/2)#

The slope of #AB# is #=(5-8)/(3-2)=(-3)/(1)=-3#

The slope of the line perpendicular to #AB# is #=1/3#

The equation of the line passing trrough #C# and perpendicular to #AB# is

#y-13/2=1/3(x-5/2)#

#y=1/3x-5/6+13/2=1/3x+17/3#

The intersection of this line with the line #y=2/7x+3# gives the center of the circle.

#1/3x+17/3=2/7x+3#

#1/3x-2/7x=3-17/3#

#1/21x=-8/3#

#x=-21*8/3=-56#

#y=2/7*(-56)+3=-13#

The center of the circle is #(-56,-13)#

The radius of the circle is

#r^2=(2+56)^2+(8+13)^2#

#=(58)^2+(21)^2#

#=3805#

The equation of the circle is

#(x+56)^2+(y+13)^2=3805#

graph{((x+56)^2+(y+13)^2-3805)(y-2/7x-3)(y-1/3x-17/3)=0 [-165.8, 101, -73.3, 60.2]}