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anonymous

  • one year ago

What is the standard form of the equation for this circle?

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @hotguy @Vocaloid @ any one who can help me

  3. anonymous
    • one year ago
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    The standard form of a circle is \((x-h)^2+(y-k)^2=r^2\) where the center of the circle is \((h, k)\) and \(r\) is the radius. Can you take it from there?

  4. anonymous
    • one year ago
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    okay so then that would mean it is (x-4)^2 + (y+5)^2 = 5.5^2

  5. anonymous
    • one year ago
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    right?

  6. anonymous
    • one year ago
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    Yup. Great job!

  7. anonymous
    • one year ago
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    thanks!

  8. anonymous
    • one year ago
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    np :)

  9. anonymous
    • one year ago
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    @Calcmathlete could you help me with another question?

  10. anonymous
    • one year ago
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    Sure

  11. anonymous
    • one year ago
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    The equation of a circle is x2 + y2 + Cx + Dy + E = 0. If the radius of the circle is decreased without changing the coordinates of the center point, how are the coefficients C, D, and E affected? i dont get how to work with circles

  12. anonymous
    • one year ago
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    Do you know how to complete the square?

  13. anonymous
    • one year ago
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    no

  14. anonymous
    • one year ago
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    Ok, completing the square is kind of a technique that is similar to factoring (somewhat). It's based on the formula \[(a\pm b)^2=a^2\pm2ab+b^2\] The idea is that you're trying to get a quadratic into a form that can become a perfect square. Before we work on your question, let's work with the idea first.\[x^2-4x+5=10\] Now, of course, the easiest thing would be to factor, but let's try completing the square instead.\[x^2-4x+\_\_=5\] You're trying to find the number that should go there to make it a perfect square trinomial.\[x^2-4x\color{red}{+4}=5\color{red}{+4}\]We added 4 on the left side to make it the perfect square trinomial (this is to make it fit the form above). However, to balance the equation, we need to add 4 to the right side as well.\[(x-2)^2=9\]That is called completing the square. Does that make sense so far?

  15. anonymous
    • one year ago
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    kind of, yes

  16. anonymous
    • one year ago
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    okay now what?

  17. anonymous
    • one year ago
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    Alright, so what we need to do is complete the square for the equation of the circle above. \[x^2+y^2+Cx+Dy+E=0\]\[(x^2+Cx+\_\_)+(y^2+Dy+\_\_)=-E\]\[\left(x^2+Cx+\left(\frac{C}{2}\right)^2\right)+\left(y^2+Dy+\left(\frac D2\right)^2\right)=-E+\left(\frac C2\right)^2+\left(\frac D2\right)^2\]\[\left(x+\frac C2\right)^2+\left(y+\frac D2\right)^2=-E+\left(\frac C2\right)^2+\left(\frac D2\right)^2\]This should look familiar from the form above.\[(x-h)^2+(y-k)^2=r^2\]where \(h=-\frac C2\), \(k=-\frac D2\), \(r^2=-E+\left(\frac C2\right)^2+\left(\frac D2\right)^2\)

  18. anonymous
    • one year ago
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    If you're not quite 100% on completing the square yet, ^ is going to be a bit hard to follow.

  19. anonymous
    • one year ago
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    okay i kinda follow its a bit hazy but its there

  20. anonymous
    • one year ago
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    Alright, I'll go over completing the square in a sec again, but to finish up your question, the variables that affect the center of the circle are C and D. Your question asked that if r decreased, how would C, D, E be affected if the center of the circle didn't change. So, since the center didn't move, C and D stay constant. However, E needs to increase for the radius to decrease. Does that make sense? I'll go back to completing the square after this.

  21. anonymous
    • one year ago
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    okay yeah that does

  22. anonymous
    • one year ago
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    the only problem is thats not an answer

  23. anonymous
    • one year ago
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    What is the answer according to your source?

  24. anonymous
    • one year ago
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  25. anonymous
    • one year ago
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    these are our options

  26. anonymous
    • one year ago
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    With the way the question is written, I don't think any of the choices are correct. Are you sure that's the exact wording of the question?

  27. anonymous
    • one year ago
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    yeah i copy and paste it

  28. anonymous
    • one year ago
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  29. anonymous
    • one year ago
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    I just double checked everything on my paper. I think their question is just straight up wrong. If it were - E or something instead of + E, then the choice would be on there. I'd just go with the last option since it's the closest one, but...wow...

  30. anonymous
    • one year ago
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    i chose e

  31. anonymous
    • one year ago
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    http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2%2B6x%2B6y%2B5%3D0 http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2%2B6x%2B6y%2B10%3D0 Also, ^ that is why I'm sure that their question is flawed.

  32. anonymous
    • one year ago
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    The +10 one is clearly smaller than the +5 one, and the centers stayed the same.

  33. anonymous
    • one year ago
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    Sorry for all the confusion.

  34. anonymous
    • one year ago
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    its okay thanks tho

  35. anonymous
    • one year ago
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    np

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