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?
Mapping a disk to a point in 3D
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0@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? – 2012-09-30
2 Answers
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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0There 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*. – 2012-09-30
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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0I think that inverse map isn't quite right, at any rate, it doesn't invert the map in the other direction. – 2012-09-30