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#

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#