Question

A circle has a center that falls on the line `y = 5/6x +8` and passes through `(4 ,8 )` and `(2 ,5 )`. What is the equation of the circle?

Answer 1

The equation of the circle is `(x-1/3)^2+(y-149/18)^2=4381/324`

Explanation:

Let the line `L` be `6y-5x=48`

Let the points `A` and `B` be

`A=(4,8)`

`B=(2,5)`

The mid-point of `AB` is

`C=((4+2)/2,(8+5)/2)=(3,13/2)`

The slope of `AB` is `m=(5-8)/(2-4)=3/2`

The slope of the line perpendicular to `AB` is

`m'=-2/3` as `mm'=-1`

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

`y-13/2=-2/3(x-3)`

`6y-39=-4x+12`

`6y+4x=51` This is line `L'`

We need the point of intersection of line `L` and `L'`

`6y-5x=48`

`6y+4x=51`

`-9x=-3`

`x=1/3`

`6y=5/3+48=149/3`

`y=149/18`

The center of the circle is `O=(1/3,149/18)`

The radius of the circle is

`r^2=((2-1/3)^2+(5-149/18)^2)`

`=25/9+3481/324`

`=4381/324`

The equation of the circle is

`(x-1/3)^2+(y-149/18)^2=4381/324`

graph{(y-5/6x-8)((x-1/3)^2+(y-149/18)^2-4381/324)=0 [-8.29, 7.505, 4.45, 12.35]}