How do you complete the square to write the equation of a sphere in standard form, and how can you find its center and radius from the equation x² + y² + z² + 2x – 6y + 8z – 1 = 0?

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To convert the given equation of the sphere into standard form and find the center and radius, we will complete the square for each of the variables (x, y, and z) in the equation:

The equation we have is:

x² + y² + z² + 2x - 6y + 8z - 1 = 0

First, we can rearrange the equation to isolate the constant term:

x² + 2x + y² - 6y + z² + 8z = 1

Now, let’s complete the square for each variable:

  1. For x: We have x² + 2x. Take half of 2 (which is 1), square it to get 1. Thus, we can write:
    2. x² + 2x = (x + 1)² - 1
  2. For y: We have y² – 6y. Take half of -6 (which is -3), and square it to get 9. So, we have:
    3. y² - 6y = (y - 3)² - 9
  3. For z: We have z² + 8z. Take half of 8 (which is 4), square it to get 16. So, we can write:
    4. z² + 8z = (z + 4)² - 16

Substituting these completed squares back into the equation gives us:

(x + 1)² - 1 + (y - 3)² - 9 + (z + 4)² - 16 = 1

Now, combine the constants on the right-hand side:

(x + 1)² + (y - 3)² + (z + 4)² - 26 = 1

To isolate the squared terms, add 26 to both sides:

(x + 1)² + (y - 3)² + (z + 4)² = 27

This is now in the standard form of a sphere, which is:

(x - h)² + (y - k)² + (z - m)² = r²

From the standard form, we can identify:

  • Center (h, k, m): (-1, 3, -4)
  • Radius (r): r = √27 = 3√3

Thus, the **center** of the sphere is at the point (-1, 3, -4), and the **radius** of the sphere is 3√3.

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