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myininaya

True/False: A graph that contains just a single point (h,k) can be written as (x-h)^2+(y-k)^2=0 which means it can be seen as a circle with radius 0. Please state what you think. Don't look it up on a website because I can do that if that is what I wanted. :)

  • one year ago
  • one year ago

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  1. ilikephysics2
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    Its true...i didnt look it up

    • one year ago
  2. klimenkov
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    It is TRUE.

    • one year ago
  3. myininaya
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    That is what I say to but people actually do take the other side on this one. If there is any reason why you think it is true, can you say why?

    • one year ago
  4. ilikephysics2
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    that is the formula for it , look in your book

    • one year ago
  5. experimentX
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    the equality only hold for (x,y) = (h,k) for any set of point other than (h, k) this relation is not valid in Real plane.

    • one year ago
  6. ilikephysics2
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    @experimentX its the formula for it man

    • one year ago
  7. experimentX
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    formula is just expression of logic.

    • one year ago
  8. ilikephysics2
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    yeah

    • one year ago
  9. CliffSedge
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    Some people don't like the idea of calling something with zero size a circle (or any other plane figure), but I don't have a problem with it. If you want to be more precise, it's the limit of a circle as its radius approaches zero.

    • one year ago
  10. CliffSedge
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    That is the standard form of an equation for a circle with r=0, so why not? I would be more general and say it's an ellipse of size zero, but I'm a dork like that.

    • one year ago
  11. myininaya
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    I think it depends on how you define a circle. I would say the radius could be greater than equal to 0. Someone told me you can actually prove that when you have (x-h)^2+(y-k)^2=0 this is a circle.

    • one year ago
  12. myininaya
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    I left out the word "or"

    • one year ago
  13. myininaya
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    or an ellipse

    • one year ago
  14. CliffSedge
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    Depends on if you want a synthetic geometry definition, analytic geometry definition, calculus definition. As far as I'm concerned, It is a circle. It's a circle with r=0.

    • one year ago
  15. CliffSedge
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    So, in more plain language (or using a geometry definition), it would be better to call it a point rather than a circle, but as long as you are clear in your definitions and can show your logic is consistent in either case, either way works.

    • one year ago
  16. Zarkon
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    if you let r=0 then you will destroy properties that all 'normal' circles have

    • one year ago
  17. myininaya
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    You put quotation marks around normal because you do see it as a circle @zarkon , but not a "normal" circle ?

    • one year ago
  18. Zarkon
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    I would call it a degenerate circle

    • one year ago
  19. CliffSedge
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    I like "degenerate."

    • one year ago
  20. CliffSedge
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    It's a shorter way of saying "circle with radius zero."

    • one year ago
  21. experimentX
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    perhaps that way we could avoid confusing the difference between these two. (x-h)^2+(y-k)^2=0 a(x-h)^2+b(y-k)^2=0

    • one year ago
  22. precal
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    but isn't a circle just defined as a center with all points on the circumference equadistant fro the center. Hmmmmm center and circumference are the same here.......

    • one year ago
  23. CliffSedge
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    I don't see why the center and circumference cannot coincide. There is nothing in the definition of a circle that forbids that.

    • one year ago
  24. precal
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    true so true........

    • one year ago
  25. Zarkon
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    that depends on what definition you use

    • one year ago
  26. Zarkon
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    I like my circles to have interiors..and the break the plane into two regions (not including the circle itself)

    • one year ago
  27. Zarkon
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    that way all my theorems hold and make sense

    • one year ago
  28. joemath314159
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    Maybe thinking of the equation as a "circle" is what makes it seem confusing. Think of it as a distance function:\[d: \mathbb{R}^2\times\mathbb{R}^2\rightarrow\mathbb{R}\]\[((x_1,y_1),(x_2,y_2))\longmapsto \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\]

    • one year ago
  29. joemath314159
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    Then in a sense, when you have:\[(x-h)^2+(y-k)^2=0\]you are saying "I want the set of all points (x,y) such that the distance between (x,y) and (h,k) is zero."

    • one year ago
  30. CliffSedge
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    @joemath314159 the equation of a circle, the distance formula, and the pythagorean theorem are all the same thing.

    • one year ago
  31. Zarkon
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    draw a tangent to a circle with radius zero

    • one year ago
  32. CliffSedge
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    @Zarkon why does a tangent need to be defined in order for it to be a circle?

    • one year ago
  33. Zarkon
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    all other circles have that property..and many other properties that are destroyed by having r=0...maybe it should have its own name...like "point"

    • one year ago
  34. CliffSedge
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    And even if there is no single unique tangent to that point, a tangent can still be drawn.

    • one year ago
  35. CliffSedge
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    Those are properties of some circles, but are they included in the definition of "circle?"

    • one year ago
  36. CliffSedge
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    Argument from personal preference isn't valid, I'd think.

    • one year ago
  37. Zarkon
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    most are theorems that require that a circle have an interior

    • one year ago
  38. CliffSedge
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    Then that's just too bad. You use those theorems when they apply. If they don't apply, you don't use the theorems. Whether or not particular theorems apply does not change the definition of "circle."

    • one year ago
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