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