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I have been told that a disk with center (a,b) and radius r can be mapped to 3D point. And the 3D point is $(a,b,a^2+b^2-r^2)$. However i do not know what is the idea behind it. How do you calculate this point and how do you prove it? Inversely if you have a 3D point can you create a disk from that point?

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You can map whatever set you want to whatever point you want. The mapping is $f(x) = p$ where $p$ is the point. That's probably not what you mean, so please tell us what you do mean. –  Robert Israel Sep 30 '12 at 4:01
    
@RobertIsrael, I mean how do we define the point in 3D? Specially why $a^2+b^2-r^2$ for the third coordinate of the point? –  kotoll Sep 30 '12 at 4:03
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A mapping is just a way of making everything in one set correspond to something in another set. Every disk has, and is determined by, a center and a radius, and the formula is just putting those three pieces of information (the two coordinates of the center, and the value of the radius) together in a triple of numbers, and every triple of numbers is a point. So, for example, the disk with center $(7,8)$ and radius 9 gets mapped to the point $(7,8,7^2+8^2-9^2)=(7,8,32)$. There is nothing to prove.

Going the other way, from $(7,8,32)$, you would say the first two entries give you the center, $(7,8)$, and then you get the radius by solving $7^2+8^2-r^2=32$ (discarding the negative value of $r$). Not every point gives you a circle, e.g., $(1,2,10)$ would require you to solve $1^2+2^2-r^2=10$, but that equation has no (real) solution.

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Then can you say change the map to i.e $(a,b,a^2+b^2+r)$ or $(a,b,a^2+b^2+r^2)$? If yes, then how do you define the mapping? I mean how do you find the good one? –  kotoll Sep 30 '12 at 4:12
    
There are many mappings, and they are all good. Now, if you have some particular use in mind for the mapping, then one mapping might be better than another for that particular use. –  Gerry Myerson Sep 30 '12 at 5:19
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If i understand your question, the answer is like this:

To determine a circle in $\mathbb{R}^{2}$, you need two things:

1- The center of the disk or equivalently two coordinates a and b.

2- The radius of the disk or equivalently a positive number r.

So using 1 and 2 you can see that given a disk on $\mathbb{R}^{2}$, you have a point on $\mathbb{R}^{3}$: (a,b,r).

Now inversely if i give you a point $(a,b,r)\in\mathbb{R}\times\mathbb{R}\times(0,\infty)$, then you have a disk on $\mathbb{R}^{2}$ with center $(a,b)$ and radius $r$.

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I think that inverse map isn't quite right, at any rate, it doesn't invert the map in the other direction. –  Gerry Myerson Sep 30 '12 at 5:20
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