Question

Find the equation of the sphere if one of its diameters has endpoints (4, 2, -9) and (6, 6, -3) which has been normalized so that the coefficient of x2 is 1.

Answer 1

`(x - 5)^2 + (y - 4)^2 + (z - 6)^2 = 14`.

(Expand to obtain an equivalent expression for the sphere:
`x^2 - 10\,x + y^2 - 8\, y + z^2 - 12\, z + 63 = 0`)

Explanation:

Apply the Pythagorean Theorem to find the distance between these two endpoints:

`\begin{aligned}&\text{Distance}\cr &= √(\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2 + \left(z_2 - z_1\right)^2) \cr &= \sqrt{(6 - 4)^2 + (6 - 2)^2 + ((-3) - (-9))^2 \cr &= √(56)}\end{aligned}`.

Since the two endpoints form a diameter of the sphere, the distance between them would be equal to the diameter of the sphere. The radius of a sphere is one-half of its diameter. In this case, that would be equal to:

`\begin{aligned} r &= (1)/(2) \, √(56) \cr &= \sqrt{\left((1)/(2)\right)^2 * 56} \cr &= \sqrt{(1)/(4) * 56} \cr &= √(14) \end{aligned}`.

In a sphere, the midpoint of every diameter would be the center of the sphere. Each component of the midpoint of a segment (such as the diameter in this question) is equal to the arithmetic mean of that component of the two endpoints. In other words, the midpoint of a segment between
`\left(x_1, \, y_1, \, z_1\right)` and
`\left(x_2, \, y_2, \, z_2\right)` would be:

`\displaystyle \left((x_1 + x_2)/(2),\, (y_1 + y_2)/(2), \, (z_1 + z_2)/(2)\right)`.

In this case, the midpoint of the diameter, which is the same as the center of the sphere, would be at:

`\begin{aligned}&\left((x_1 + x_2)/(2),\, (y_1 + y_2)/(2), \, (z_1 + z_2)/(2)\right) \cr &= \left((4 + 6)/(2),\, (2 + 6)/(2), \, ((-9) + (-3))/(2)\right) \cr &= (5,\, 4\, -6)\end{aligned}`.

The equation for a sphere of radius
`r` and center
`\left(x_0,\, y_0,\, z_0\right)` would be:

`\left(x - x_0\right)^2 + \left(y - y_0\right)^2 + \left(z - z_0\right)^2 = r^2`.

In this case, the equation would be:

`\left(x - 5\right)^2 + \left(y - 4\right)^2 + \left(z - (-6)\right)^2 = \left(√(56)\right)^2`.

Simplify to obtain:

`\left(x - 5\right)^2 + \left(y - 4\right)^2 + \left(z + 6\right)^2 = 56`.

Expand the squares and simplify to obtain:

`x^2 - 10\,x + y^2 - 8\, y + z^2 - 12\, z + 63 = 0`.