Question

A circle has a center that falls on the line `y = 1/3x +7` and passes through `( 3 ,1 )` and `(6 ,9 )`. What is the equation of the circle?

Answer 1

`color(red)((x--15/34)^2+(y-233/34)^2=26645/578)`

Explanation:

For the first point `(3, 1)` we can set up one equation

`(x-h)^2+(y-k)^2=r^2`

`(3-h)^2+(1-k)^2=r^2`

For the second point `(6, 9)` we can set up another equation

`(x-h)^2+(y-k)^2=r^2`

`(6-h)^2+(9-k)^2=r^2`
~~~~~~~~~~~~~~~~~~~~~~~~~~
`r^2=r^2`
`(3-h)^2+(1-k)^2=(6-h)^2+(9-k)^2`

After simplification, we now have an equation

`6h+16k=107`

Also from the line `y=x/3+7` which contains the point `(h, k)`, we have

`k=h/3+7` or `h-3k=-21`

Simultaneous solution of the system
`h-3k=-21`
`6h+16k=107`

results to a center `(h, k)=(-15/34, 233/34)`

the radius can be computed now using center `(h, k)=(-15/34, 233/34)` and `(6-h)^2+(9-k)^2=r^2`

`(6-h)^2+(9-k)^2=r^2`
`r^2=(6--15/34)^2+(9-233/34)^2`
`r^2=26645/578`

We can now form the equation of the circle

`(x-h)^2+(y-k)^2=r^2`

`color(red)((x--15/34)^2+(y-233/34)^2=26645/578)`

Kindly see the graph for visual inspection

God bless....I hope the explanation is useful.