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Naudiapie

  • one year ago

What is the equation of a circle that passes through (2, 0) and has its center at M(4, 0)?

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  1. Naudiapie
    • one year ago
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    please help!!!!

  2. anonymous
    • one year ago
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    standard form of circle: (x - h)^2 + (y - k)^2 = r^2center is (h , k) and radius is rfor your circle, the center (h , k) is (1 , 2)so it will look like: (x - 1)^2 + (y - 2)^2 = r^2to find the radius squared, plug in the (x,y) point you know is on the circle -- (3 , 4)(3 - 1)^2 + (4 - 2)^2 = r^22^2 + 2^2 = 8 = r^2

  3. jdoe0001
    • one year ago
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    |dw:1437264018410:dw|

  4. Naudiapie
    • one year ago
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    ???

  5. anonymous
    • one year ago
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    Ok, so we know the standard form of a circle is (x - h) ² + (y - k) ² = r ² and the center is (h , k) and radius is r. Does this make sense so far?

  6. Naudiapie
    • one year ago
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    yes

  7. Naudiapie
    • one year ago
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    !!!

  8. jim_thompson5910
    • one year ago
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    How far is it from (4,0) to (2,0) ?

  9. Naudiapie
    • one year ago
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    (2,0) ?? i dont know xD

  10. jim_thompson5910
    • one year ago
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    |dw:1437263870646:dw|

  11. jim_thompson5910
    • one year ago
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    do you see how they are 2 units apart ? |dw:1437263878771:dw|

  12. Naudiapie
    • one year ago
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    yes i see that

  13. anonymous
    • one year ago
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    I'll brb, I have to finish some of my tests

  14. jim_thompson5910
    • one year ago
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    |dw:1437264011320:dw|

  15. Naudiapie
    • one year ago
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  16. Naudiapie
    • one year ago
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    it makes no sense to me

  17. jim_thompson5910
    • one year ago
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    The general circle equation is this |dw:1437264186407:dw|

  18. jim_thompson5910
    • one year ago
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    (h,k) is the center r is the radius

  19. jim_thompson5910
    • one year ago
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    we're told that the center is (4,0) so (h,k) = (4,0) meaning that h = 4, k = 0 |dw:1437264245046:dw|

  20. jim_thompson5910
    • one year ago
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    |dw:1437264266650:dw|

  21. jim_thompson5910
    • one year ago
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    r = 2 was just found and shown above |dw:1437264297767:dw|

  22. jim_thompson5910
    • one year ago
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    |dw:1437264320072:dw|

  23. jim_thompson5910
    • one year ago
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    also notice how \(\Large (y-0)^2\) turned into \(\Large y^2\)

  24. Naudiapie
    • one year ago
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    how did you get 2 squared, can you explain that?

  25. Naudiapie
    • one year ago
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    that's what i don't get

  26. jim_thompson5910
    • one year ago
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    well the right side of the general circle equation has \(\Large r^2\) I replaced r with 2 because the radius is r = 2

  27. jim_thompson5910
    • one year ago
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    r = 2 \[\Large r^2 = 2^2\]

  28. jim_thompson5910
    • one year ago
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    |dw:1437264515515:dw|

  29. Naudiapie
    • one year ago
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    ohh i see okay

  30. Naudiapie
    • one year ago
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    so x - 4squared + ysquared = 2squared (aka 4)

  31. jim_thompson5910
    • one year ago
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    yes \[\Large (x-4)^2 + y^2 = 4\]

  32. jim_thompson5910
    • one year ago
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    make sure you use parenthesis don't say x-4^2 say (x-4)^2

  33. Naudiapie
    • one year ago
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    okay thanks a lot , i get it now

  34. jim_thompson5910
    • one year ago
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    4^2 means "4 squared"

  35. Naudiapie
    • one year ago
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    thank you lots

  36. anonymous
    • one year ago
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    You could also have used the equation of the circle to figure out the radius algebraicly. We knew \[ (x-4)^2+y^2=r^2 \]And we knew \((x,y)=(2,0)\) lie in the circle, so \[ (2-4)^2+0^2=r^2 \\ (-2)^2 = r^2 \implies |r| = 2 \]But since we know \(r\) is positive for it to be a circle, we pick \(r=2\).

  37. Naudiapie
    • one year ago
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    :o

  38. jim_thompson5910
    • one year ago
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    you're welcome

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