Question

How do you write an equation of a circle whose diameter of the circle are (-2,6) and (-8,10)?

Answer 1

Please see the explanation.

`(x - -5)^2 + (y - 8)^2 = (sqrt(13))^2`

Explanation:

The general 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.

To go from `(-2, 6) " to " (-8, 10)` one must move to the left 6 and up 4, therefore, the center must be to the left 3 and up 2; the point `(-5, 8)`. Substitute the center into equation [1]:

`(x - -5)^2 + (y - 8)^2 = r^2" [2]"`

Substitute the point `(-2,6)` into equation [2]:

`(-2 - -5)^2 + (6 - 8)^2 = r^2`

Solve for r:

`(3)^2 + (-2)^2 = r^2`

`13 = r^2`

`r = sqrt(13)`

Substitute `r = sqrt(13)` into equation [2]:

`(x - -5)^2 + (y - 8)^2 = (sqrt(13))^2`

This is the equation of the circle.