Find the equation of the sphere if one of it's diameters has the endpoints (2,8,4) and (6,1,3)?

Question

I'm more interested in how it's done than in the answer.

Answer 1

First, find the center, which is the average of the coordinates of the two given ends of a diameter.
C(4, 9/2, 7/2).
Then find the length of the diameter with the distance formula.
d = √(2-6)^2 + (8-1)^2 + (4-3)^2 = √((-4)^2 + 7^2 + 1^2) =
= √(16+49+1) = √66
the radius is half the diameter:
r = d/2 = √66/2
r^2 = 66/4 = 33/2
the general equation of a sphere is:
(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2 where (a,b,c) is the center and r is the radius.
So the equation is
(x-4)^2 + (y-9/2)^2 + (z-7/2)^2 = 33/2

Answer 2

Find the length of the diameter and divide it by 2 Find the midpoint Plug those values into this formula: (x - h)^2 + (y - i)^2 + (z - j)^2 = r^2 d^2 = (2 - 0)^2 + (12 - (-4))^2 + (-3 - 7)^2 d^2 = 2^2 + 16^2 + (-10)^2 d^2 = 4 + 256 + 100 d^2 = 360 d^2 = 6 * 6 * 10 d = 6 * sqrt(10) r = d/2 r = 3 * sqrt(10) r^2 = 9 * 10 r^2 = 90 h = (0 + 2) / 2 = 2/2 = 1 i = (-4 + 12) / 2 = 8/2 = 4 k = (7 + -3) / 2 = 4/2 = 2 (x - 1)^2 + (y - 4)^2 + (z - 2)^2 = 90 There you go.

Answer 3

You can write the equation of a sphere as:

(x - a)^2 + (y - b)^2 + (z - c)^2 = R^2

Where a, b and c are the offsets to the center of the sphere. If at the origin they are all 0.

R is the radius of the sphere
D = 2R = SQRT((6-2)^2 + (1 - 8)^2 + (3 - 4)^2)
R = SQRT(16 + 49 + 1)/2 = SQRT(66)/2

The center of the sphere is the half way point between the two given points. Solve for this point.

Enough?