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Find the equation of a sphere if one of its diameters has endpoints: (-14. -3, -6) and (-4, 7, 4) Note that you must move everything to the left hand side of the equation and that we desire the coefficients of the quadratic terms to be 1.

Sagot :

okpalawalter8

Answer:

[tex]x^2+y^2+z^2-18x-4y+2z-21=0[/tex]

Step-by-step explanation:

From the question we are told that:

Diameters has endpoints: [tex](-14. -3, -6) & (-4, 7, 4)[/tex]

Generally the equation for Center of The sphere is mathematically given by

 [tex]C=(\frac{-14+(-4)}{2},\frac{-3+(7)}{2},\frac{-6+(4)}{2})[/tex]

 [tex]C=(9,2,-1)[/tex]

Generally the equation for Radius of the sphere is mathematically given by

 [tex]R=\sqrt{(9-2)^2+(2-9)^2+(-1-2)^2}[/tex]

 [tex]R=\sqrt{107}[/tex]

Therefore the Equation of the Sphere is

 [tex](x-9)^2+(y-2)^2+(z+1)^2=107[/tex]

 [tex](x^2-18x+81)+(y^2-4y+4+(z^2+2z+1))=107[/tex]

 [tex]x^2+y^2+z^2-18x-4y+2z-21=0[/tex]

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