Question

Center and Radius of a Sphere Show that the equation represents a sphere, and find its center and radius.

$x^2+y^2+z^2=12x+2y$.

Short answer

The given equation $x^2+y^2+z^2=12x+2y$ is the equation of sphere with center $(6,1,0)$ and radius $\sqrt{37}$

Step-by-step solution

Step 1. Given information.

Given: $x^2+y^2+z^2=12x+2y$

Step 2. We have to show that the equation represents a sphere, and find its center and radius.

We will complete the squares in x,y and z terms to rewrite the given equation in the form of an equation of sphere.

The given equation is

$\begin{array}{c}{x}^2+y^2+z^2=12x+2y\\ x^2+y^2+z^2-12x-2y=0\\ =\left(x^2-12x+36\right)+\left(y^2-2y+1\right)+\left(z^2\right)=0+36+1\\ ={(x-6)}^2+{(y-1)}^2+(z+0)=37\end{array}$

add samenumbers both side.
Comparing this with standard equation of sphere, we can see that the center is $(6,1,0)$and the radius is $\sqrt{37}$