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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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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).
Posted by sciencesolve on December 3, 2011 at 4:25 PM (Answer #1)
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