# How do I find the equation of the sphere of radius 2 centered at the origin?

Sep 10, 2015

Use Pythagoras theorem twice to derive a distance formula, hence the equation:

${x}^{2} + {y}^{2} + {z}^{2} = {2}^{2}$

#### Explanation:

The distance of a point $\left(x , y , z\right)$ from $\left(0 , 0 , 0\right)$ is

$\sqrt{{x}^{2} + {y}^{2} + {z}^{2}}$

To see this you can use Pythagoras twice:

The points $\left(0 , 0 , 0\right)$, $\left(x , 0 , 0\right)$ and $\left(x , y , 0\right)$ form the vertices of a right-angled triangle with sides of length $x$, $y$ and $\sqrt{{x}^{2} + {y}^{2}}$.

Then the points $\left(0 , 0 , 0\right)$, $\left(x , y , 0\right)$ and $\left(x , y , z\right)$ form the vertices of a right angled triangle with sides of length $\sqrt{{x}^{2} + {y}^{2}}$, $z$ and

$\sqrt{{\left(\sqrt{{x}^{2} + {y}^{2}}\right)}^{2} + {z}^{2}} = \sqrt{{x}^{2} + {y}^{2} + {z}^{2}}$

So we can write the equation of our sphere as:

$\sqrt{{x}^{2} + {y}^{2} + {z}^{2}} = 2$

or

${x}^{2} + {y}^{2} + {z}^{2} = {2}^{2}$