Find an equation of the sphere that passes through the point (4 3 -1) and has center (3 8 1)

Solution:

The standard form of the equation of a sphere is

(x - a)² + ( y - b)² + ( z - c)² = r²

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

From the question we should first find the radius

r = √[(x - a)² + ( y - b)² + ( z - c)²]

Point = (x, y, z) = (4, 3, -1)

Center = (a, b, c) = (3, 8, 1)

Substituting it in the formula

r = √[(4 - 3)² + (3 - 8)² + (-1 - 1)²]

By further calculation

r = √[1 + 25 + 4]

So we get

r = √30

Substituting it in the standard form

(x - 3)² + ( y - 8)² + ( z - 1)² = √30²

(x - 3)² + ( y - 8)² + ( z - 1)² = 30

⇒ x² - 6x + 9 + y² - 16y + 64 + z² - 2z +1 = 30 

⇒ x² + y² + z² - 6x -16y - 2z + 9 + 64 + 1 = 30

⇒ x² + y² + z² - 6x - 16y - 2z + 44 = 0

Therefore, the equation of the sphere is x² + y² + z² - 6x - 16y - 2z + 44 = 0

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Find an equation of the sphere that passes through the point (4 3 -1) and has center (3 8 1)

Summary:

The equation of the sphere that passes through the point (4 3 −1) and has center (3 8 1) is x² + y² + z² - 6x - 16y - 2z + 44 = 0