Question

Determine the equation of a circle whose diameter has the endpoints (–1, 2) and (7, –4).

Answer 1

(x - 3)² + (y + 1)² = 25

Explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r is the radius

the centre C is at the midpoint of the endpoints of the diameter

using the midpoint formula

C = (
`(x_(1)+x_(2) )/(2)` ,
`(y_(1)+y_(2) )/(2)` )

with (x₁, y₁ ) = (- 1, 2 ) and (x₂, y₂ ) = (7, - 4 )

C = (
`(-1+7)/(2)` ,
`(2-4)/(2)` ) = (
`(6)/(2)` ,
`(-2)/(2)` ) = (3, - 1 )

the radius r is the distance from the centre to either of the 2 endpoints

using the distance formula

r =
`\sqrt{(x_(2)-x_(1))^2+(y_(2)-y_(1))^2 }`

with (x₁, y₁ ) = C (3, - 1 ) and (x₂, y₂ ) = (7, - 4 )

r =
`√((7-3)^2+(-4-(-1))^2)`

=
`√(4^2+(-4+1)^2)`

=
`√(16+(-3)^2)`

=
`√(16+9)`

=
`√(25)`

= 5

Then equation of circle is

(x - 3)² + (y-(- 1))² = 5² , that is

(x - 3)² + (y +1)² = 25