Question

if p=(3,1) and Q=(-3,-7), find the equation of the circle that has segment PQ as the diameter (x-{?})^2+(y-{?})^2={?}

Answer 1

x² + (y + 3)² = 25

Explanation:

the centre (C) of the circle is at the midpoint of the diameter.

using the midpoint formula

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

with (x₁, y₁ ) = P (3, 1 ) and (x₂, y₂ ) = Q (- 3, - 7 )

C = (
`(3-3)/(2)` ,
`(1-7)/(2)` ) = (
`(0)/(2)` ,
`(-6)/(2)` ) = (0, - 3 )

the radius r is the distance from the centre to either P or Q

using the distance formula

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

with (x₁, y₁ ) = C (0, - 3 ) and (x₂, y₂ ) = P (3, 1 )

r =
`√((3-0)^2+(1-(-3)^2)`

=
`√(3^2+(1+3)^2)`

=
`√(3^2+4^2)`

=
`√(9+16)`

=
`√(25)`

= 5

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

here (h, k ) = (0, - 3 ) and r = 5 , then

(x - 0 )² + (y - (- 3) )² = 5² , that is

x² + (y + 3)² = 25