Question

A circle has a center that falls on the line `y = 7/6x +1` and passes through `(1 ,4 )` and `(8 ,5 )`. What is the equation of the circle?

Answer 1

The equation of the circle is `(x-210/49)^2+(y-294/49)^2=35525/2401`

Explanation:

Let `C` be the mid point of `A=(1,4)` and `B=(8,5)`

`C=((1+8)/2,(4+5)/2)=(9/2,9/2)`

The slope of `AB` is `=(5-4)/(8-1)=1/7`

The slope of the line perpendicular to `AB` is `=-7`

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

`y-9/2=-7(x-9/2)`

`y=-7x+63/2+9/2=-7x+36`

The intersection of this line with the line `y=7/6x+1` gives the center of the circle.

`7/6x+1=-7x+36`

`7/6x+7x=36-1`

`49/6x=35`

`x=6*35/49=210/49`

`y=-7*210/49+36=294/49`

The center of the circle is `(210/49,294/49)`

The radius of the circle is

`r^2=(1-210/49)^2+(4-294/49)^2`

`=(161/49)^2+(98/49)^2`

`=35525/2401`

The equation of the circle is

`(x-210/49)^2+(y-294/49)^2=35525/2401`
graph{((x-210/49)^2+(y-294/49)^2-35525/2401)(y-7/6x-1)=0 [-11.24, 14.07, -1.11, 11.55]}