How do you write the equation of the sphere 2x^2 + 2y^2 + 2z^2 – 8x – 24z + 1 in standard form and find its center and radius?

To rewrite the equation of the sphere in standard form, we start with the given equation:

2x2 + 2y2 + 2z2 – 8x – 24z + 1 = 0

First, we can simplify the equation by dividing everything by 2:

x2 + y2 + z2 – 4x – 12z + rac{1}{2} = 0

Next, we rearrange the equation:

x2 – 4x + y2 + z2 – 12z = - rac{1}{2}

Now we will complete the square for the x and z terms:

For the x terms: x2 – 4x can be rewritten as (x – 2)2 – 4 (since (-4/2)2 = 4).

For the z terms: z2 – 12z can be rewritten as (z – 6)2 – 36 (since (-12/2)2 = 36).

Putting it all together, we have:

(x – 2)2 – 4 + y2 + (z – 6)2 – 36 = - rac{1}{2}

Simplifying gives:

(x – 2)2 + y2 + (z – 6)2 = 39.5

This is now in the standard form of a sphere’s equation: (x – h)2 + (y – k)2 + (z – l)2 = r2, where (h, k, l) is the center and r is the radius.

The center is:

(h, k, l) = (2, 0, 6)

The radius (r) can be calculated as the square root of 39.5:

r = √39.5 ≈ 6.293

In summary:

  • Center: (2, 0, 6)
  • Radius: ≈ 6.293

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