Question

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?

Answer 1

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`