A circle has a center that falls on the line #y = 6/7x +7 # and passes through # ( 7 ,8 )# and #(3 ,9 )#. What is the equation of the circle?

1 Answer
Nov 8, 2016

graph{(6/7x+7-y)(y-17/2-4(x-5))(y-9+1/4(x-3))((x-7)^2+(y-8)^2-0.04)((x-3)^2+(y-9)^2-0.04)((x-259/44)^2+(y-265/22)^2-0.04)((x-259/44)^2+(y-265/22)^2-34085/44^2)=0 [-5, 19, 5, 17]}

#(x-259/44)^2+(y-265/22)^2=34085/44^2#

Explanation:

Call the points #A# and #B#
Get the direction of #bar(AB)#

#m_(bar(AB))=(8-9)/7-3=-1/4#

The direction of #bar(AB)# axis is

#m_(bar(AB)|--)=4#

Let #M# be the medium point of #bar(AB)#, then

#M=((3+7)/2;(9+8)/2)=(5;17/2)#

The equation of #bar(AB)# axis is

#y-17/2=4(x-5)#

The center #C# of the circle must beleng to #bar(AB)# axis so

#{(y_C=4(x_C-5)+17/2),(y_C=6/7x_C+7):}#

Solving the system we obtain

#C=(259/44;265/22)#

and for the radius

#r=sqrt((x_A-x_C)^2+(y_A-y_C)^2)=sqrt((259/44-3)^2+(265/22-9)^2)=sqrt(34085)/44#