Question

A circle has a center that falls on the line `y = 3/2x +6` and passes through `(1 ,2 )` and `(6 ,1 )`. What is the equation of the circle?

Answer 1

The equation of the circle is `(x-44/7)^2+(y-108/7)^2=208.3`

Explanation:

Let the center of the circle be `(a,b)` and radius r
Then the equation of the circle is
`(x-a)^2+(y-b)^2=r^2`

Then, `(1-a)^2+(2-b)^2=r^2`
and `(6-a)^2+(1-b)^2=r^2`

Equating the equations
`(1-a)^2+(2-b)^2=(6-a)^2+(1-b)^2`
developing,
`1-2a+a^2+4-4b+b^2=36-12a+a^2+1-2b+b^2`

`5-2a-4b=37-12a-2b`
`10a-2b=32`
`5a-b=16` this is equation 1
Also, `y=(3x)/2+6`
So `b=(3a)/2+6`
Solving for a and be, we get `a=44/7` and `b=108/7`
We calculate the radius `r^2=(6-44/7)^2+(1-108/7)^2`

`r^2=(2/7)^2+(108/7)^2` `=>` `r=14.43`

Equation of circle is
`(x-44/7)^2+(y-108/7)^2=208.3`