Would someone mind checking this for me?

Find the center and radius of the sphere given by the equation
$$3x ^{2}+3y^{2}+3z^{2}+2y-2z=9$$
$$x^{2}+y^{2}+\frac{2}{3}y+z^{2}-\frac{2}{3}z=3\ (regroup\ and\ divide\ by\ 3)$$
$$x^{2}+y^{2}+\frac{2}{3}y+\frac{1}{9}+z^{2}-\frac{2}{3}z+\frac{1}{9}=3-\frac{1}{9}-\frac{1}{9}\ (Complete\ the\ \Square)$$
$$x^{2}+(y+\frac{1}{3})^{2}+(z-\frac{1}{3})^{2}=\frac{25}{9}$$
$$Center\ (0,-\frac{1}{3},\frac{1}{3}),\ radius\ \frac{5}{3}$$

Question

Would someone mind checking this for me?

Find the center and radius of the sphere given by the equation
$$3x ^{2}+3y^{2}+3z^{2}+2y-2z=9$$
$$x^{2}+y^{2}+\frac{2}{3}y+z^{2}-\frac{2}{3}z=3\ (regroup\ and\ divide\ by\ 3)$$
$$x^{2}+y^{2}+\frac{2}{3}y+\frac{1}{9}+z^{2}-\frac{2}{3}z+\frac{1}{9}=3-\frac{1}{9}-\frac{1}{9}\ (Complete\ the\ \Square)$$
$$x^{2}+(y+\frac{1}{3})^{2}+(z-\frac{1}{3})^{2}=\frac{25}{9}$$
$$Center\ (0,-\frac{1}{3},\frac{1}{3}),\ radius\ \frac{5}{3}$$

amistre64 · Answer

http://www.wolframalpha.com/input/?i=3x%5E2+%2B3y%5E2+%2B3z%5E2+%2B2y%E2%88%922z%3D9+

amistre64 · Answer

centers good, radius seems off a bit

dumbcow · Answer

looks pretty good.

the only thing i see is when you complete the square you should have added 2/9 to right side

amistre64 · Answer

i tend to divide off the 3 after i complete the square parts to avoid fractions

chriss · Answer

right... I subtracted it instead of added... thanks!

chriss · Answer

so my radius should be 
$$\frac{\sqrt{29}}{3}$$

amistre64 · Answer

yep

chriss · Answer

awesome.  thanks to you both :)