Question

A circle has a center that falls on the line `y = 2/7x +7` and passes through `( 3 ,4 )` and `(6 ,1 )`. What is the equation of the circle?

Answer 1

`(x - 63/5)^2 + (y - 53/5)^2 = ((3sqrt377)/5)^2`

Explanation:

The general Cartesian form for the equation of a circle is:

`(x-h)^2 + (y-k)^2 = r^2" [1]"`

Substitute the point `(3,4)` into equation [1]:

`(3-h)^2 + (4-k)^2 = r^2" [2]"`

Substitute the point `(6,1)` into equation [1]:

`(6-h)^2 + (1-k)^2 = r^2" [3]"`

Expand the squares of equation [2] and [3]:

`9-6h+h^2 + 16-8k+k^2 = r^2" [4]"`
`36-12h+h^2 + 1-2k+k^2 = r^2" [5]"`

Subtract equation [5] from equation [4]:

`color(white)(....)9-6h+h^2 + 16-8k+k^2 = r^2" [4]"`
`ul(-36+12h-h^2 - 1+2k-k^2 = -r^2)" [5]"`
`-27+6h+0h^2+15-6k+0k^2= 0r^2`

Remove all of the 0 terms and move the constant to the right:

`6h - 6k = 12" [6]"`

Evaluate the given equation, `y = 2/7x +7`, at the center `(h,k)`:

`k = 2/7h+7" [7]"`

Substitute equation [7] into equation [6]:

`6h - 6(2/7h+7) = 12`

Solve for h:

`42h -12h- 294 = 84`

`30h = 378`

`h = 63/5`

Use equation [7] to find the value of k:

`k = 2/7(63/5)+7`

`k = 53/5`

Use equation [2] to find the value of r:

`(3-63/5)^2 + (4-53/5)^2 = r^2`

`(15/5-63/5)^2 + (20/5-53/5)^2 = r^2`

`(-48/5)^2 + (-33/5)^2 = 3393/25 = r^2`

`r = (3sqrt377)/5`

Use equation [1] to write the equation:

`(x - 63/5)^2 + (y - 53/5)^2 = ((3sqrt377)/5)^2`