Question

How do you find the center, radius, and equation of a circle with : (1,-2) and (7,6) as the diameter's endpoints?

Answer 1

`(x-4)^2 +(y-2)^2 = 25`

Explanation:

The centre of the circle `c` is at the midpoint of the diameter, and the radius `r`is equal to half the diameter. We therefore need the distance between the two points.

Using Pythagoras to calculate the length of the diameter `xy` gives
`(xy)^2 = (7-1)^2 + (6-(-2))^2`
`(xy)^2 = 36 +64 =100`

`:. yx =sqrt(100) = 10`

The radius is therefore `5`

`c` is at the midpoint of `xy`. Its `x` coordinate `h` is therefore midway between `x` and `z`, and its `y` coordinate `k` is midway between `z` and `y`.
`h = 1 + (7-1)/2 = 4`
`k = -2 + (6-(-2))/2 = 2`

The equation of the circle is therefore `(x-4)^2 +(y-2)^2 = 25`