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What is the center of the sphere x^2+y^2+z^2+2x-8y-4z-15=0

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lxsptter | Student, Undergraduate | Valedictorian

Posted December 3, 2011 at 4:08 PM via web

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What is the center of the sphere x^2+y^2+z^2+2x-8y-4z-15=0

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted December 3, 2011 at 4:25 PM (Answer #1)

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Equation of a sphere in `R^3`  is:

`(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2 = r^2`

The point `(x_0,y_0,z_0) ` is the center of the sphere and r is the radius of the sphere.

You should create groups of terms of the same variable, such that

`(x^2 + 2x) + (y^2 - 8y) + (z^2 - 4z) - 15 = 0`

Complete the perfect squares:

`(x^2 + 2x + 1) + (y^2 - 8y + 16) + (z^2 - 4z + 4) - 15-1 - 16 - 4= 0`

Use special products:

`(x+1)^2 + (y-4)^2 + (z-2)^2 - 36 = 0`

Add 36 both sides:

`(x+1)^2 + (y-4)^2 + (z-2)^2 = 36`

`(x+1)^2 + (y-4)^2 + (z-2)^2 = 6^2`

Comparing the standard equation of the sphere to the equation above you'll have:

the center of the sphere is (-1 ; 4 ; 2) 

the radius of the sphere r = 6

The coordinates of the center of the sphere: (-1 ; 4 ; 2).

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