Question

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

Answer 1

`(x - 27)^2 + (y - 25)^2 = (3sqrt(113))^2`

Explanation:

The general form of the equation of a circle is:

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

where `(x, y)` is any point on the circle, `(h, k)` is the center, and r is the radius.

Use equation [1] to write two equations, using the given points:

`(3 - h)^2 + (4 - k)^2 = r^2" [2]"`
`(6 - h)^2 + (1 - k)^2 = r^2" [3]"`

Temporarily eliminate r by setting the left side of equation [2] equal to the left side of equation [3]:

`(3 - h)^2 + (4 - k)^2 = (6 - h)^2 + (1 - k)^2`

Expand the squares, using the pattern `(a - b)^2 = a^2 - 2ab + b^2`:

`9 - 6h + h^2 + 16 - 8k + k^2 = 36 - 12h + h^2 + 1 - 2k + k^2`

The square terms cancel:

`9 - 6h + 16 - 8k = 36 - 12h + 1 - 2k`

Collect all of the constant terms into a single term on the right:

`-6h - 8k = -12h - 2k + 12`

Collect all of the h terms into a single term on the right:

`- 8k = -6h - 2k + 12`

Collect all of the k terms into a single term on the left:

`-6k = -6h + 12`

Divide both sides of the equation by -6:

`k = h - 2" [4]"`

Evaluate the given equation at the point `(h, k)`

`k = 2/3h + 7" [5]"`

Set the right side of equation [4] equal to the right side of equation [5]:

`h - 2 = 2/3h + 7`

`1/3h = 9`

`h = 27`

Substitute 27 for h in equation [5]:

`k = 2/3(27) + 7`

`k = 18 + 7`

`k = 25`

Substitute the center, `(27,25)`, into equation [2] and solve for r:

`(3 - 27)^2 + (4 - 25)^2 = r^2`

`(-24)^2 + (-21)^2 = r^2`

`r^2 = 1017`

`r = sqrt(1017) = 3sqrt(113)`

Substitute the center, `(27,25)`, and the radius, `3sqrt(113)` into equation [1]:

`(x - 27)^2 + (y - 25)^2 = (3sqrt(113))^2`

This is the equation of the circle.