Find the center and radius of the sphere.
-$$3 x ^ { 2 } + 3 y ^ { 2 } + 3 z ^ { 2 } - 2 x + 2 y = 9$$
A) $\left( \frac { 1 } { 3 } , - \frac { 1 } { 3 } , 0 \right) , \mathrm { a } = \frac { \sqrt { 29 } } { 3 }$
B) $\left( - \frac { 1 } { 9 } , \frac { 1 } { 9 } , 0 \right) , a = \frac { \sqrt { 29 } } { 9 }$
C) $\left( - \frac { 1 } { 3 } , \frac { 1 } { 3 } , 0 \right) , a = \frac { 29 } { 9 }$
D) $\left( - \frac { 1 } { 3 } , \frac { 1 } { 3 } , 0 \right) , \mathrm { a } = \frac { \sqrt { 29 } } { 3 }$

Question

Quizplus · Accepted Answer

The equation of the sphere is in the form $$(x-h)^2+(y-k)^2+(z-l)^2=r^2$$where (h,k,l) is the center and r is the radius.Completing the square for x and y, we get$$3x^2-2x+3y^2+2y=9$$$$3(x^2-\frac{2}{3}x+\frac{1}{9})+3(y^2+\frac{2}{3}y+\frac{1}{9})=9+3(\frac{1}{9})+3(\frac{1}{9})$$$$3(x-\frac{1}{3})^2+3(y+\frac{1}{3})^2=9+1+1$$$$(x-\frac{1}{3})^2+(y+\frac{1}{3})^2=11/3$$$$(x-\frac{1}{3})^2+(y+\frac{1}{3})^2+(\frac{z}{1})^2=(\frac{\sqrt{29}}{3})^2$$So the center is $$\left( \frac { 1 } { 3 } , - \frac { 1 } { 3 } , 0 \right)$$ and the radius is $$\frac { \sqrt { 29 } } { 3 }$$

Find the Center and Radius of the Sphere 3x2+3y2+3z2−2x+2y=93 x ^ { 2 } + 3 y ^ { 2 } + 3 z ^ { 2 } - 2 x + 2 y = 93x2+3y2+3z2−2x+2y=9

Find the Center and Radius of the Sphere $3 x ^ { 2 } + 3 y ^ { 2 } + 3 z ^ { 2 } - 2 x + 2 y = 9$