Question

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

Answer 1

Equation of the circle
`(x-13/2)^2+(y-47/2)^2=1546/4`

Explanation:

Let the center `(h,k)`

From the given line `y=3x+4`
we can write `k=3h+4` because `(h, k)` is on the line

Using the given points (4, 4) and (9, 4)

radius=radius

`sqrt((4-h)^2+(4-k)^2)=sqrt((9-h)^2+(4-k)^2)`

also

`(4-h)^2+(4-k)^2=(9-h)^2+(4-k)^2`

`16-8h+h^2=81-18h+h^2`

`18h-8h=81-16`
`10h=65`
`h=13/2`

Using now the equation `k=3h+4`

`k=3(13/2)+4`

`k=39/2+4`

`k=(39+8)/2`

`k=47/2`

the center of the circle
`(h, k)=(13/2, 47/2)`

compute the radius `r` now

`r^2=(4-h)^2+(4-k)^2`

`r^2=(4-13/2)^2+(4-47/2)^2`

`r^2=((8-13)/2)^2+((8-47)/2)^2`

`r^2=(-5/2)^2+(-39/2)^2`

`r^2=1546/4`

We can now obtain the equation of the circle

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

`(x-13/2)^2+(y-47/2)^2=1546/4`

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