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hba

Equation of a circle with ends of diameter at (-3,2) and (5,-6) is :

  • one year ago
  • one year ago

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  1. hba
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    @hartnn @waterineyes I have no clue about this one........

    • one year ago
  2. hartnn
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    that is simple.

    • one year ago
  3. hartnn
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    mid-point is centre

    • one year ago
  4. waterineyes
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    Find the distance between and divide it by 2 to find radius..

    • one year ago
  5. hartnn
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    distance between the two points = diameter =2r

    • one year ago
  6. waterineyes
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    Do you know how to find midpoint or not??

    • one year ago
  7. hba
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    I know nothing lol but i am going to give you a tutorial on Conic section soon promise :D

    • one year ago
  8. hba
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    @waterineyes noooooooo

    • one year ago
  9. hartnn
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    u don't know how to find mid-point ? thats news... just take the average of co-ordinates.

    • one year ago
  10. hba
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    x+y/2 but x1 or x2

    • one year ago
  11. hartnn
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    for x-coordinate of midpoint, u take average of x-coordinates of endpoints

    • one year ago
  12. hartnn
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    (-3+5)/2

    • one year ago
  13. waterineyes
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    Sorry net stopped working..

    • one year ago
  14. waterineyes
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    And I think your mind too.. @hba

    • one year ago
  15. waterineyes
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    \[Midpoint(x,y) = (\frac{x_1 +x_2}{2}, \frac{y_1+y_2}{2})\]

    • one year ago
  16. hba
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    @waterineyes Yeah right lol and [Math processing error ] Anyways @hartnn I remember (x,y)=(x1+x2/2,y1+y2/2)

    • one year ago
  17. hba
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    @waterineyes I cannot see latex

    • one year ago
  18. hartnn
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    so whats the midpoint ? centre

    • one year ago
  19. hba
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    (x,y)=(1,-2)

    • one year ago
  20. hartnn
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    thats correct centre

    • one year ago
  21. hartnn
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    now find radius

    • one year ago
  22. waterineyes
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    Just refresh your page then..

    • one year ago
  23. hba
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    So i use the old radius formula ?

    • one year ago
  24. hba
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    \[r=\sqrt{g^2+f^2-c}\]

    • one year ago
  25. hba
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    @waterineyes @hartnn Do you have a better way ?

    • one year ago
  26. hartnn
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    not that, can u find distance between two points ?

    • one year ago
  27. waterineyes
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    What are you doing man?? Use distance formula..

    • one year ago
  28. hba
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    Yeah

    • one year ago
  29. hartnn
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    Distance between points (x1,y1) and (x2,y2) is \(\huge d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

    • one year ago
  30. waterineyes
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    \[Distance = \sqrt{(x_2-x_1)^2 - (y_2-y_1)^2}\]

    • one year ago
  31. hba
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    \[d=\sqrt{(x2-x1)^2+(y2-y1)^2}\]

    • one year ago
  32. hba
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    lol

    • one year ago
  33. waterineyes
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    Written three times but used not atleast once.. ha ha ha..

    • one year ago
  34. hartnn
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    lol :P

    • one year ago
  35. hba
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    \[d=8\sqrt{2}\]

    • one year ago
  36. hba
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    what next ?

    • one year ago
  37. hartnn
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    radius = d/2

    • one year ago
  38. hba
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    @hartnn ?

    • one year ago
  39. hartnn
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    commented

    • one year ago
  40. waterineyes
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    This you have got is length of Diameter.. You have to find radius so you will half it..

    • one year ago
  41. hba
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    r=4root(2)

    • one year ago
  42. hartnn
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    yup... u know what to do from now....

    • one year ago
  43. waterineyes
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    |dw:1353175013593:dw|

    • one year ago
  44. hba
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    (x-a)^2+(y-b)^2=r^2

    • one year ago
  45. waterineyes
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    Yep Go Ahead..

    • one year ago
  46. hba
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    x^2+y^2-2x+4y-27=0

    • one year ago
  47. waterineyes
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    Yep..

    • one year ago
  48. waterineyes
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    1 + 4 = 32 implies -27 on LHS..

    • one year ago
  49. hba
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    Thanks a lot @hartnn and @waterineyes

    • one year ago
  50. waterineyes
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    You are welcome dear..

    • one year ago
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