Question

Find the equation of a circle given only its diameter endpoints: (-2,1) and (6,-7)

Answer 1

(x-2)^2+(y+3)^2=32
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Answer 2

(x - 2)² + (y + 3)² = 32

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 of the circle is at the midpoint of the diameter

Calculate the centre (x, y ) using the midpoint formula

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

with (x₁, y₁ ) = (- 2, 1) and (x₂, y₂ ) = (6, - 7)

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

The radius is the distance from the centre to either of the endpoints

Calculate the radius using the distance formula

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

with (x₁, y₁ ) = (2, - 3) and (x₂, y₂ ) = (- 2, 1)

r =
`√((-2-2)^2+(1-(-3))^2)`

=
`\sqrt{(-4)^2+(1+3)^2`

=
`√(16+4^2)`

=
`√(16+16)`

=
`√(32)`

Then equation of circle is

(x - 2)² + (y - (- 3) )² = (
`√(32)` )² , that is

(x - 2)² + (y + 3)² = 32