Question

A circle has a center that falls on the line `y = 5/3x +1` and passes through `(8 ,2 )` and `(3 ,1 )`. What is the equation of the circle?

Answer 1

The equation of the circle is `(x-21/5)^2+(y-8)^2=1261/25`

Explanation:

Let the points `A=(8,2)` and `B=(3,1)`

The midpoint of `AB` is `C=((8+3)/2, (1+2)/2)=(11/2,3/2)`

The slope of `AB` is `m=(1-2)/(3-8)=-1/-5=1/5`

The slope of the line perpendicular to `AB` is `m'=-5`, as

`mm'=-1`

The equation of the line, passing through `C` and perpendicular to `AB` is `L'`

`y-3/2=-5(x-11/2)`

`y-3/2=-5x+55/2`

`y=-5x+55/2+3/2=-5x+29`

Let the line `y=5/3x+1` be `L`

The center of the circle `O` lies at the intersection of `L` and `L'`

We have 2 equations

`y=5/3x+1` and `y=-5x+29`

So,

`5/3x+1=-5x+29`

`5/3x+5x=29-1=28`

`20/3x=28`

`x=28/20*3=21/5`

`y=5/3*21/5+1=8`

The center of the circle is `O=(21/5,8)`

The radius of the circle is

`r^2=(21/5-3)^2+(8-1)^2=36/25+49=1261/25`

The equation of the circle is

`(x-21/5)^2+(y-8)^2=1261/25` graph{(y-5/3x-1)(y+5x-29)((x-21/5)^2+(y-8)^2-1261/25)=0 [-18.56, 21.97, -2.56, 17.72]}