# What is the center of the sphere with the equation (x-2)^2+(y+3)^2+(z-6)^2=36?

Sep 1, 2015

$\left(2 , - 3 , 6\right)$

#### Explanation:

The center of any sphere written in the form:
$\left(x - a\right) + \left(y - b\right) + \left(z - c\right) = {r}^{2} \text{ }$, is $\left(a , b , c\right)$

${\left(x - 2\right)}^{2} + {\left(y + 3\right)}^{2} + {\left(z - 6\right)}^{2} = 36$ can be written as:

${\left(x - 2\right)}^{2} + {\left(y - \left(- 3\right)\right)}^{2} + {\left(z - 6\right)}^{2} = {\left(6\right)}^{2}$

hence center is at $\textcolor{b l u e}{\left(2 \text{,"-3",} 6\right)}$

Sep 1, 2015

The centre has coordinates: $\left(+ 2 , - 3 , + 6\right)$
1) The coordinates of the centre are values next to $x , y$ (and $z$) with changed signs,