Is a Circle with Radius 0 Real?

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A circle with a radius of 0 is considered a "degenerate" circle, represented as a single point rather than a traditional circle. While it technically does not fit the standard definition of a circle, it can still be conceptualized as such in certain contexts. A negative radius leads to an empty set, which some may view as an extreme form of degeneration. The discussion also touches on the idea of circles in complex space, where negative or complex radii could be considered, but this complicates the traditional understanding of circles. Ultimately, a circle is best defined with a positive, non-zero radius for clarity and consistency.
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if it is 0 ...

hey all ...
if we have the circle equation and we found the radius ... if it is positive the circle would off course be real ... but if it was 0, do we say that we don't have a circle or the circle is a dot ?
cheers
abc
 
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The graph is a single point. Technically, that's NOT a circle but you can consider it a "degenerate" circle.

(and if r< 0, your graph is the empty set- some people consider that a REALLY degenerate circle!)
 
"Technically",the equation would be (for one with the center in the origin)
x^{2}+y^{2}=R^{2} and the radius being negative would not mean anything wrong.The pairs (x,y) satisfying it would be the same.As for "zero",well,that should be seen as the "degenerate" case...

The wrong part with negative radius would come from the definition of the circle,which uses the word "distance",and the definition of distance for a metric space (in this case R^{2}) which is an application to R_{+}\cup \{0\}..

Daniel.
 
One could argue that the circle in question exists in complex space where either or both of x and y are complex.

Then again, if R^2 is negative the radius is also complex. does this invalidate the nature of a circle or is it reasonable?
 
No,it does not invalidate the "nature" of the circle,it just pushes our power of understanding...The abstractization is sometimes unappropriate.The circle MUST have a real positive nonzero radius.And it's preferable to have as a one-dimensional object...

Daniel.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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