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anonymous

  • one year ago

Using the following equation, find the center and radius of the circle. You must show all work and calculations to receive credit. x2 + 2x + y2 + 4y = 20 I am completely lost on how to do this!!! I will fan and medal please someone help!

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  1. anonymous
    • one year ago
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    @ganeshie8 @zepdrix @mathmate

  2. mathmate
    • one year ago
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    Complete the square for x and y, then you'll get the equation of the circle with centre (h,k) and radius r as in (x-h)^2+(y-k)^2 = r^2

  3. mathmate
    • one year ago
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    @cheetah21 Can you do that or do you need further help?

  4. anonymous
    • one year ago
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    I do need further help..sorry im just really bad with math :/

  5. mathmate
    • one year ago
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    Start with the given equation, we need to transform it to the standard form of the circle, then we can identify (by comparison) h,k and r.

  6. anonymous
    • one year ago
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    Okay how do we do that ?

  7. mathmate
    • one year ago
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    The given equation is: x2 + 2x + y2 + 4y = 20 After adding a 1 and a 4 on both sides, we put the equation as (x^2+2x+1) + (y^2+2(2y)+4) = 20 +1 +4 It's still the same equation, but what are in the parentheses are perfect squares. Can you factor the two expressions in parentheses?

  8. anonymous
    • one year ago
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    x2 + 2x + y2 + 4y = 20 (x2 + 2x) + (y2 + 4y) = 20

  9. mathmate
    • one year ago
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    I already put them in perfect squares, you don't need to simplify them. In fact: (x^2+2x+1) + (y^2+2(2y)+4) = 20 +1 +4 is the same as: (x+1)^2 + (y+2)^2 = 25 You can check to make sure by expanding the squares.

  10. anonymous
    • one year ago
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    Oh okay im sorry , i was confused i guess :/ so now do we need to solve in the parentheses or what ?

  11. mathmate
    • one year ago
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    Next step is easy: you compare (x+1)^2 + (y+2)^2 = 5^2 with the standard form of the circle with centre (h,k) and radius r (x-h)^2+(y-k)^2 = r^2 and decide what the values of h, k and r are.

  12. anonymous
    • one year ago
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    (X-h)^2 + (y - K)^2 = r^2 (x-h)^2+(y-k)^2 = r^2 is that what it should look like right now ?? and how do we figure out values ?

  13. mathmate
    • one year ago
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    Compare term by term, for example, (x+1)^2 = (x-h)^2, what would be the value of h to make the right hand side (x+1)^2?

  14. anonymous
    • one year ago
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    Would that be 1 also?

  15. mathmate
    • one year ago
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    If h=1, then we have (x+1)^2 on the left hand side, and (x-1)^2 , so the two sides don't equal. You need to find another value of h.

  16. anonymous
    • one year ago
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    so then that would be 2 ?

  17. mathmate
    • one year ago
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    If h=2, then we have (x+1)^2 on the left hand side, and (x-2)^2 , so the two sides don't equal. Remember we are looking for h in: (x+1)^2 = (x-h)^2

  18. anonymous
    • one year ago
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    oh 4 ?

  19. mathmate
    • one year ago
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    (x+1)^2 = (x-4)^2 (are they equal?)

  20. anonymous
    • one year ago
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    (5x)=(25x)

  21. anonymous
    • one year ago
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    no wait typo

  22. mathmate
    • one year ago
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    There is a positive sign on the left, and a negative sign on the right, so h must be negative to make the two sides match, remembering -(-1)=+1. so putting h=-1 will make (x+1)^2 = (x-(-1))^2=(x+1)^2 or both sides equal.

  23. mathmate
    • one year ago
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    Can you try finding k from: (y+2)^2 = (y-k)^2 ?

  24. anonymous
    • one year ago
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    okay so then -4

  25. mathmate
    • one year ago
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    You would take the square root on both sides, so (y+2) = y-k, can you solve for k?

  26. anonymous
    • one year ago
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    (3) = yk 3yk Im sorry if thats a dumb answer i just dont know

  27. mathmate
    • one year ago
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    From (y+2) = y-k, subtract y from both sides, so y-y+2 = y-y-k or 2=-k, multiply by -1 to get -2=k, so we have h=-1, k=-2, and r=5 From the standard equation of a circle, (x-h)^2+(y-k)^2=r^2 we therefore found centre = (h,k)=-1,-2, and radius = r =5

  28. anonymous
    • one year ago
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    wait so thats the answer to the eqaution ? oh okay , well if i put all the stuff we have come up with together can you tell me if its in the correct order?

  29. mathmate
    • one year ago
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    put it up, and perhaps someone else will look at it for you. I have to go now, sorry! :(

  30. anonymous
    • one year ago
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    Its okay, thank you for all your help!

  31. mathmate
    • one year ago
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    no problem! :)

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