A circle has a center that falls on the line `y = 7/9x +7` and passes through `( 2 ,5 )` and `(5 ,1 )`. What is the equation of the circle?

The general equation of a circle is:

`r² = (x - h)² + (y - k)²`

where r is the radius and (h, k) is the center point.

We can use the above to write two equations, using the given points:

`r² = (2 - h)² + (5 - k)²`
`r² = (5 - h)² + (1 - k)²`

Because `r² = r²` we can set the right sides equal:

`(2 - h)² + (5 - k)² = (5 - h)² + (1 - k)²`

I will use the pattern `(a - b)² = a² - 2ab + b²` to expand the squares:

`4 - 4h + h² + 25 - 10k + k² = 25 - 10h + h² + 1 - 2k + k²`

Combine like terms:

`3 + 6h = 8k`

Flip and divide both sides by 8:

`k = 3/4h + 3/8`

Substitute (h, k) into the given equation of the line on which the center lies:

`k = 7/9h + 7`

Because `k = k`, we can use the right sides of both linear equations for the center to solve for h:

`7/9h + 7 = 3/4h + 3/8`

Multiply both sides of the equation by 72:

`56h + 504 = 54h + 27`

`2h = -477`

`h = -477/2`

`k = -357/2`

Using the point (2, 5) and the center `(-477/2, -357/2)` we compute the square of the radius:

`r² = (2 - -477/2)² + (5 - -357/2)²`

`r² = 91512.5`

The equation of the circle is:

`91512.5 = (x - -477/2)² + (y - -357/2)²`