Question

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

Answer 1

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 `(x, y, z)` from `(0, 0, 0)` is

`sqrt(x^2+y^2+z^2)`

To see this you can use Pythagoras twice:

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

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

`sqrt((sqrt(x^2+y^2))^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`