A circle has a center that falls on the line #y = 8/7x +2 # and passes through # ( 2 ,1 )# and #(3 ,6 )#. What is the equation of the circle?

1 Answer
Nov 6, 2016

#C(70/47,174/47)# radius#=root2(16705/2209)#
Equation of circumference is #(x-70/47)^2+(y-174/47)^2=16705/2209#

Explanation:

#A(2,1)# and #B(3,6)# a point #C# on the line #y=8/7x+2# has coordinates #C(x,8/7x+2)#.

The centre #C# can be found by imposing that #AC=BC#

#root2((2-x)^2+(1-8/7x-2)^2)=root2((3-x)^2+(6-8/7x-2)^2)#
By squaring boh members we get

#4-4x+x^2+1+16/7x+64/49x^2=9-6x+x^2+16-64/7x+64/49x^2#
that properly manipulated becomes
#16/7x-4x+64/7x+6x=25-5# from which
#x=70/47# and consequently
#y=8/7*70/47+2=80/47+2=174/47#.
These are the coordinates of the centre #C#
The squared radius is #R^2=(2-70/47)^2+(1-174/47)^2=16705/2209#