Center and Radius of simple sphere Equation

Question

Find Radius and Center Point of This Sphere

$$x^2+y^2+z^2=6z$$ please i want to understand where this sphere located exactly

The Answer in the book is

$M(0,0,3), r=3$

can anyone check from this maybe wrong in the book?

Answer 1

You can write the equation in the form $$x^2+y^2+(z-1)^2=1.$$ Can you work out the centre and radius of this sphere?

Answer 2

write your equation in this form $$x^2+y^2+z^2-2z+1=1$$

Answer 3

Your equation is equivalent to $$x^2+y^2+(z-1)^2=1$$ So the center is the point with coordinates $M(0,0,1)$ and the radius is $1$, because this equation describes the set of all points $P(x,y,z)$ that satisfy $$d(P,M)=1$$ where $d$ is the Euclidean metric.

Answer 4

$$\begin{align}x^2+y^2+z^2&=6z\\x^2+y^2+z^2-6z&=0\\x^2+y^2+(z^2-6z+9)&=9\\x^2+y^2+(z-3)^2&=3^2\end{align}$$

Hence, the center is at $(0,0,3)$ with radius $3$