Question

A circle has a center that falls on the line `y = 8/9x +4` and passes through `( 3 ,1 )` and `(5 ,7 )`. What is the equation of the circle?

Answer 1

The equation of the circle is
`(x-1.091)^2+(y-4.97)^2=4.41^2`

Explanation:

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

The circle passes through the two points. So
`(3-a)^2+(1-b)^2=r^2`
and
`(5-a)^2+(7-b)^2=r^2`

We can equalize the two equations

`(3-a)^2+(1-b)^2=(5-a)^2+(7-b)^2`
Developing and simplifying
`9-6a+a^2+1-2b+b^2=25-10a+a^2+49-14b+b^2`

`10-6a-2b=74-10a-14b`

`12b=-4a+64` `=>` `3b=-a+16`, this equation 1

We obtain the equation 2 from the equation of the line
`b=(8a)/9+4`

Solving for `a` and `b`, we get `a=1.091` and `b=4.97`
and the radius of the circle is

`r^2=(3-1.091)^2+(1-4.97)^2`
`r^2=1.909^2+3.97^2`

`r=4.41`
So the equation of the circle is
`(x-1.091)^2+(y-4.97)^2=4.41^2`