Question

If the endpoints of the diameter of a circle are (6, -2)and (-6 ,2)what is the standard form equation of the circle

Answer 1

`x^2 + y^2 - 40 = 0`

Explanation:

Given

`(h_1,k_1) = (6,-2)`

`(h_2,k_2) = (-6,2)`

Required

Determine the equation of the circle

First, calculate the midpoint of the given coordinates:

`(h,k) = (1)/(2)(h_1+h_2,k_1+k_2)`

`(h,k) = (1)/(2)(6-6,-2+2)`

`(h,k) = (1)/(2)(0,0)`

`(h,k) = (0,0)`

Next, calculate the radius.

This is the distance between
`(0,0)` calculated above and any of
`(6,-2)`
`(-6,2)`

Using:

`(0,0)` and
`(-6,2)`, the radius is:

`r = √((0 - (-6))^2 + (0-2)^2)`

`r = √(40)`

The equation is then calculated using:

`(x - h)^2 + (y - k)^2 = r^2`

Where:

`r = √(40)` and
`(h,k) = (0,0)`

`(x - 0)^2 + (y - 0)^2 = (√(40))^2`

`(x )^2 + (y)^2 = 40`

`x^2 + y^2 = 40`

`x^2 + y^2 - 40 = 0`