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What is the radius of the sphere $x^{2}+y^{2}+z^{2}-6 x+8 y-10 z$ $+1=0 ?$
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The correct answer is:
7
The equation of sphere is $x^{2}+y^{2}+z^{2}-6 x+8 y-10 z+1=0$
Comparing the equation with general form of sphere, $x^{2}+y^{2}+z^{2}+2 u x+2 v y+2 w z-d-0$
we get, $u=\frac{-6}{2}=-3, v=\frac{8}{2}=4, w=\frac{-10}{2}=5, d=1$
Radius $=\sqrt{u^{2}+v^{2}+w^{2}-d}$
$=\sqrt{(-3)^{2}+(4)^{2}+(5)^{2}-1}$
$=\sqrt{9+16+25-1}=\sqrt{49}=7$
Comparing the equation with general form of sphere, $x^{2}+y^{2}+z^{2}+2 u x+2 v y+2 w z-d-0$
we get, $u=\frac{-6}{2}=-3, v=\frac{8}{2}=4, w=\frac{-10}{2}=5, d=1$
Radius $=\sqrt{u^{2}+v^{2}+w^{2}-d}$
$=\sqrt{(-3)^{2}+(4)^{2}+(5)^{2}-1}$
$=\sqrt{9+16+25-1}=\sqrt{49}=7$
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