find the standard form of the equation for the circle described below?

Question

Endpoints of a diameter are (9,2) and (-9,-12)?

Answer 1

The mid point of the diameter is center of the circle.

mid point is given by

xm = (x1+x2)/2

ym = (y1+y2)/2

here x1 = 9, y1 = 2

x2 = -9 , y2 = -12

xm = (9-9)/2 = 0

ym = (2 -12)/2 = -5

so center is at (0,-5)

now find out the length of radius by distance formula between any one of the end points and center

r = sqrt[[(x1-xm)^2 + (y1-ym)^2]

r = sqrt[(9 - 0)^2 + (2 -(-5)^2]

r = sqrt(81 + 49) = sqrt(130)

r^2 = 130

the equation of circle in standard form is

(x - h)^2 + (y - k)^2 = r^2

where (h, k) center coordinates

here h = 0 and k = -5

x^2 + (y+5)^2 = 130

Answer 2

If the circle's center is NOT at the origin, the equation of the circle is
(x-a)^2 + (y-b)^2 = k^2, where k is the radius.
From geometry, the diameter is sqrt(18^2+14^2)
or 2sqrt(130). The radius is sqrt(130). The center is located halfway between the extreme points of the diameter, which is at 0,-5. So the circle has the equation x^2 + (y+5)^2 = 130

Answer 3

diameter goes through the center and is halfway. So finding the midpoint of the the points given will give you the circle's center (h,k)

midpoint = ((9 + -9)/2, (2 + -12)/2)
mdpt = (0, -5)

Then the radius is the distance from the center to one endpoint
Using the distance formula with (0, -5) and (9,2)

r= sqrt((9 - 0)^2 + (2 - -5)^2)
r = sqrt(81 + 49)
r = sqrt(130)

Circle: (x - h)^2 + (y - k)^2 = r^2
(x -0)^2 + (y --5)^2 = sqrt(130)^2

x^2 + (y + 5)^2 = 130

Answer 4

Find the center point :
Center Point = ( (9-9) / 2 , (2-12) / 2 ) = ( 0 , -5 )

Find the radius :
Radius = √ [ (9 + 9)² + (2 + 12)² ] = √ (324 + 196) = √520

Now frame the equation,

(x - Cx)² + (y - Cy)² = r²

x² + (y + 5)² = 520