Question

Find the standard form for the equation of a circle (x-h)^2 + (y-k)^2 = r^2 with a diameter that has endpoints of (-2, 8) and (2, 3).

Answer 1

In the standard equation
\[(x-h)^2+(y-k)^2=r^2\]
(h, k) is the center of the circle
r= radius
we are given the end points of the diameter
A(2, -8) and B(2, 3)
diameter obviously passes through the center and we can find radius as well if we are given the diameter.
@ballerinamd could you find the length of the radius?

Answer 2

It just occurred to me...

1.) Center point: (h,k) = (0,5.5)

[The coordinate (0,5.5) equivalent to the midpoint of the diameter I derived from the endpoints.]

Equation of a circle: (x-h)^2+(y-k)^2=r^2
(x-0)^2 + (y-5.5)^2 = r^2

Substitute one of the end-points into the equation to derive the radius:
End-point: (2,3)
Substitution: (2-0)^2+(3-5.5)^2 = r^2
r^2 = 10.25
r = 3.201

Radius = 3.201

(Thank You!)