Would someone mind checking this for me? Find the center and radius of the sphere given by the equation \[3x ^{2}+3y^{2}+3z^{2}+2y-2z=9\] \[x^{2}+y^{2}+\frac{2}{3}y+z^{2}-\frac{2}{3}z=3\ (regroup\ and\ divide\ by\ 3)\] \[x^{2}+y^{2}+\frac{2}{3}y+\frac{1}{9}+z^{2}-\frac{2}{3}z+\frac{1}{9}=3-\frac{1}{9}-\frac{1}{9}\ (Complete\ the\ \Square)\] \[x^{2}+(y+\frac{1}{3})^{2}+(z-\frac{1}{3})^{2}=\frac{25}{9}\] \[Center\ (0,-\frac{1}{3},\frac{1}{3}),\ radius\ \frac{5}{3}\]

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Would someone mind checking this for me? Find the center and radius of the sphere given by the equation \[3x ^{2}+3y^{2}+3z^{2}+2y-2z=9\] \[x^{2}+y^{2}+\frac{2}{3}y+z^{2}-\frac{2}{3}z=3\ (regroup\ and\ divide\ by\ 3)\] \[x^{2}+y^{2}+\frac{2}{3}y+\frac{1}{9}+z^{2}-\frac{2}{3}z+\frac{1}{9}=3-\frac{1}{9}-\frac{1}{9}\ (Complete\ the\ \Square)\] \[x^{2}+(y+\frac{1}{3})^{2}+(z-\frac{1}{3})^{2}=\frac{25}{9}\] \[Center\ (0,-\frac{1}{3},\frac{1}{3}),\ radius\ \frac{5}{3}\]

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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http://www.wolframalpha.com/input/?i=3x%5E2+%2B3y%5E2+%2B3z%5E2+%2B2y%E2%88%922z%3D9+
centers good, radius seems off a bit
looks pretty good. the only thing i see is when you complete the square you should have added 2/9 to right side

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i tend to divide off the 3 after i complete the square parts to avoid fractions
right... I subtracted it instead of added... thanks!
so my radius should be \[\frac{\sqrt{29}}{3}\]
yep
awesome. thanks to you both :)

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