Question

A circle has a center that falls on the line `y = 7/2x +3` and passes through `(1 ,2 )` and `(8 ,1 )`. What is the equation of the circle?

Answer 1

`7x^2 - 132x +7y^2 - 504y + 1105 = 0`

Explanation:

Point A `(1,2)` and point B `(8,1)` must be the same distance (one radius) from the centre of the circle
This is lies on the line of points (L) that are all equi-distant from A and B

the formula for calculating the distance (d) between two points (from pythagorus) is `d^2 = (x_2-x_1)^2+(y_2-y_1)^2`

substitute in what we know for point A and an arbitrary point on L
`d^2 = (x-1)^2 + (y-2)^2`
substitute in what we know for point B and an arbitrary point on L
`d^2 = (x-8)^2 + (y-1)^2`

Therefore
`(x-1)^2 + (y-2)^2 = (x-8)^2 + (y-1)^2`

Expand the brackets
`x^2-2x+1 +y^2-4y+4 = x^2 -16x+64 + y^2 -2y +1`

Simplify
`2x + 4y = 16x + 2y - 60`
`2y = 14x - 60`
`y = 7x -30`

the centre point lies on the line `y = 7x - 30` (the set of points equi-distant from A and B)
and on the line `y = 7x/2 + 3` (given)

solve where these two lines cross to find the centre of the circle
`7x - 30 = 7x/2 + 3`
`14x -60 = 7x +6`
`7x = 66`
`x = 66/7`

substitute into `y = 7x/2 + 3`
`y = 7*66/(7*2) + 3 = 36`

The centre of the circle is at `(66/7, 36)`

the squared radius of the circle can now be calculated as
`r^2 = (66/7 - 1)^2 + (36-2)^2`
`r^2 = (59/7)^2 + 34^2`

The general formula for a circle or radius `r` is
`(x – h)^2 + (y – k)^2 = r^2` with the centre at h,k

We now know `h`, `k` and `r^2` and can substitute them into the general equation for the circle
`(x – 66/7)^2 + (y – 36)^2 = (59/7)^2 + 1156`

expand the brackets
`x^2 - 132x/7 + 4356/49 + y^2 -72y + 1296 = 3481/49 +1156`
and simplify
`7x^2-132x+7y^2-504y=3481/7 -7*1296 -4356/7+7*1156`

`7x^2 - 132x +7y^2 - 504y + 1105 = 0`