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

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

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Answer 1 · 2015-09-10T17:34:50

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

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

Explanation:

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

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

To see this you can use Pythagoras twice:

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

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

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

So we can write the equation of our sphere as:

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

or

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

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

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