Question

A circle has a center that falls on the line `y = 2/9x +4` and passes through `( 3 ,1 )` and `(5 ,7 )`. What is the equation of the circle?

Answer 1

`(x - 12/5)^2 + (y - 68/15)^2 = (17sqrt(10)/15)^2`

Explanation:

The equation for a circle:

`(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] and the two given points to write two equations:

`(3 - h)^2 + (1 - k)^2 = r^2" [2]"`
`(5 - h)^2 + (7 - k)^2 = r^2" [3]"`

Substitute h for x and k for y in the given equation:

`k = 2/9h + 4`

Multiply both sides of the equation by 9:

`9k = 2h + 36`

Write in standard form:

`2h - 9k = -36" [4]"`

Because `r = r` we can set the left side of equation [2] equal to the left side of equation [3]:

`(3 - h)^2 + (1 - k)^2 = (5 - h)^2 + (7 - k)^2`

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

`9 - 6h + h^2 + 1 - 2k + k^2 = 25 - 10h + h^2 + 49 - 14k + k^2`

The `h^2 and k^2` terms cancel:

`9 - 6h + 1 - 2k = 25 - 10h + 49 - 14k`

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

`-6h - 2k = -10h - 14k + 64`

Combine the h terms and the k terms on the left:

`4h + 12k = 64`

Divide both sides of the equation by 4:

`h + 3k = 16" [5]"`

Here are the two equations that define the values of h and k:

`2h - 9k = -36" [4]"`
`h + 3k = 16" [5]"`

Multiply equation [5] by 3 and add to equation [4]:

`5h = 12`

`h = 12/5`

Substitute `12/5` for h in equation [4]:

`2(12/5) - 9k = -36`

`9k = 36 + 24/5`

`k = 68/15`

To find the value of r, substitute the value for h and k into equation [3]:

`(5 - 12/5)^2 + (7 - 68/15)^2 = r^2"`

`(5 - 36/15)^2 + (7 - 68/15)^2 = r^2"`

`r = 17sqrt(10)/15`

Substitute these values into equation [1]:

`(x - 12/5)^2 + (y - 68/15)^2 = (17sqrt(10)/15)^2`