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rainbow22

  • 4 years ago

Center of a circle with this equation: y^2-6x-8y+20=-x^2

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  1. Mertsj
    • 4 years ago
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    Get the variables on the left and the constant term on the right. complete the square twice and write the equation in the form: \[(x-h)^2+(y-k)^2=r^2\]

  2. rainbow22
    • 4 years ago
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    I did but I end up with what I believe is the wrong answer. Okay, so obviously it would be (x-4)^2+(y-3)^3=-20

  3. rainbow22
    • 4 years ago
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    I am confused as to what to do with the neg.

  4. Mertsj
    • 4 years ago
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    What negative?

  5. rainbow22
    • 4 years ago
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    The neg in front of the twenty

  6. Callisto
    • 4 years ago
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    y^2-6x-8y+20=-x^2 x^2 +y^2-6x-8y+20= 0 centre = (-D/2 , -E/2) in the form x^2 +y^2+Dx+Ey+F= 0 Therefore centre = ( -(-6/2) , -(-8/2) ) =...... Can you do it now?

  7. Mertsj
    • 4 years ago
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    Nothing. It will be gone because you will be adding numbers to the right and the left sides.

  8. Mertsj
    • 4 years ago
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    Ok. callisto wants to help you now. Good bye and good luck.

  9. rainbow22
    • 4 years ago
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    center is therefore.. (-3,-4) right?

  10. Callisto
    • 4 years ago
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    No... the sign is not correct

  11. rainbow22
    • 4 years ago
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    sooo it's (3,4).

  12. rainbow22
    • 4 years ago
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    So basically it's still the same center as it would be without the negatives...?

  13. Callisto
    • 4 years ago
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    Yes

  14. rainbow22
    • 4 years ago
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    Okay. Also, what the radius would be sqrtroot of 20.. right?

  15. rainbow22
    • 4 years ago
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    Sorry, but just for conformation.. It doesn't matter if there is a negative for the 'r', the center would still be the same.

  16. Callisto
    • 4 years ago
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    Nope.. \[r = \sqrt { (D/2)^2 + (E/2)^2 -F}\]

  17. Callisto
    • 4 years ago
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    D=-6, E=-8 , F=20 in your case. Put the numbers into it and solve, what would you get for r?

  18. rainbow22
    • 4 years ago
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    squareroot of 5...

  19. Callisto
    • 4 years ago
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    It should be that ...

  20. rainbow22
    • 4 years ago
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    That.. doesn't make sense though. The radius is typically sqroot of \[(x-h)^2 + (y-k)^2=r^2\]

  21. Callisto
    • 4 years ago
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    So, let's do the completing square once y^2-6x-8y+20=-x^2 x^2 + y^2-6x-8y+20= 0 x^2 -6x +y^2-8y+20 =0 (x^2 -6x +9 -9) +(y^2-8y +16 -16)+20 =0 (x^2 -6x +9) +(y^2-8y +16) -9 -16 +20 =0 (x-3)^2 + (y-4)^2 - 5 =0 (x-3)^2 + (y-4)^2 = 5 (x-3)^2 + (y-4)^2 =[ sqrt (5) ]^2 r = sqrt 5 It DOES make sense

  22. Callisto
    • 4 years ago
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    Here is the proof...

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