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The radius of the sphere $\quad[2015-I I]$ $3 x^{2}+3 y^{2}+3 z^{2}-8 x+4 y+8 z-15=0$ is
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3
$3 x^{2}+3 y^{2}+3 z^{2}-8 y+4 y+8 z-15=0$
$\Rightarrow x^{2}+y^{2}+z^{2}-\frac{8}{3} x+\frac{4}{3} y+\frac{8}{3} z-5=0$
$\Rightarrow\left(x-\frac{4}{3}\right)^{2}-\frac{16}{9}+\left(y+\frac{2}{3}\right)^{2}-\frac{4}{9}+\left(z+\frac{4}{3}\right)^{2}-\frac{16}{9}-5=0$
$\Rightarrow\left(x-\frac{4}{3}\right)^{2}+\left(y+\frac{2}{3}\right)^{2}+\left(z+\frac{4}{3}\right)^{2}=(3)^{2}$
So radius is 3 .
$\Rightarrow x^{2}+y^{2}+z^{2}-\frac{8}{3} x+\frac{4}{3} y+\frac{8}{3} z-5=0$
$\Rightarrow\left(x-\frac{4}{3}\right)^{2}-\frac{16}{9}+\left(y+\frac{2}{3}\right)^{2}-\frac{4}{9}+\left(z+\frac{4}{3}\right)^{2}-\frac{16}{9}-5=0$
$\Rightarrow\left(x-\frac{4}{3}\right)^{2}+\left(y+\frac{2}{3}\right)^{2}+\left(z+\frac{4}{3}\right)^{2}=(3)^{2}$
So radius is 3 .
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