Question

Find the standard form of the equation of the circle with endpoints of a diameter at the points (3,8) and (-7,2)

Answer 1

`(x+2)^(2)`+
`(y-5)^(2)`= 34

The equation of a circle in standard form is

`(x-a)^(2)`+
`(y-b)^(2)`=
`r^(2)`

where (a , b) are the coordinates of the centre and r is the radius

The centre is at the midpoint of the endpoints and the radius is the distance from the centre to either of the 2 endpoints

Using the midpoint formula

midpoint = [
`(1)/(2)`(x
`x_(1)`+
`x_(2)`,
`(1)/(2)`(
`y_(1)`+
`y_(2)`

where (
`x_(1)`,
`y_(1)`=(3,8) and (
`x_(2)`,
`y_(2)`=(-7,2)

centre = (
`(1)/(2)`(3-7),
`(1)/(2)`(8+2)) = (-2,5)

Calculate r using the distance formula

r = √(
`x_(2)`-
`x_(1)`)²+(
`y_(2)`-
`y_(1)`)²)

= √((3+2)²+(8-5)²) = √(25+9) = √34 ⇒ r² =(√34)² = 34

equation of circle is : (x+2)²+(y-5)² = 34