Write the equation of the sphere in standard form 2x² + 2y² + 2z² = 4x - 16z + 1, Find its center and radius?

Solution:

The equation of the sphere in standard form is (x - a)² + (y - b)² + (z - c)² = r²

Where C(a, b, c) is the center and r is the radius

Given: 2x² + 2y² + 2z² = 4x - 16z + 1

Divide both sides by 2

x² + y² + z² = 2x - 8z + 1/2

(x² - 2x) + y² + (z² + 8z) = 1/2

(x² - 2 x (1) x (x) + 12) - 12 + y² + (z² + 2 x 4 x (z) + 42) - 42 = 1/2

(x - 1)² + y² + (z + 4)² = 1/2 + 1 + 16

Taking LCM

(x - 1)² + y² + (z + 4)² = (1 + 2 + 32) / 2

(x - 1)² + y² + (z + 4)² = 35/2

(x - 1)² + y² + (z + 4)² = √(35/2)²

Therefore, the equation of the sphere in standard form is (x - 1)² + y² + (z + 4)² = √(35/2)².

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Write the equation of the sphere in standard form 2x² + 2y² + 2z² = 4x - 16z + 1, Find its center and radius?

Summary:

The equation of the sphere in standard form 2x² + 2y² + 2z² = 4x - 16z + 1 is (x - 1)² + y² + (z + 4)² = √(35/2)², its center = (1, 0, -4) and radius is √(35/2).