Question

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

Answer 1

The equatio of the circle is `(x-56/15)^2+(y-31/3)^2=87.65`

Explanation:

Let the center of the cicle be `(a,b)`
As the line passes through the center `b=(5a)/2+1`
The equation of the circle is `(x-a)^2+(y-b)^2=r^2`
where `r` is the radius
As the circle passes through `(8,2)` and (3,1)
we get, `(8-a)^2+(2-b)^2=r^2`
and `(3-a)^2+(1-b)^2=r^2`
Equating `(8-a)^2+(2-b)^2=(3-a)^2+(1-b)^2`
Developing
`64-16a+a^2+4-4b+b^2=9-6a+a^2+1-2b+b^2`

`64-16a+4-4b=9-6a-2b+1`

`68-16a-4b=10-6a-2b`

`58=10a+2b`

`5a+b=29` we compare this to the first equation
`b=(5a)/2+1`
`29-5a=(5a)/2+1`
`15a/2=28`
`a=2*28/15=56/15`
and `b=5/2*56/15+1=31/3`
So the radius is `r^2=(3-56/15)^2+(1-31/3)^2=11^2/15^2+28^2/9=87.65`
and `r=9.36`