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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 − 6y = 26

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  1. jdoe0001
    • one year ago
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    do you know what a "perfect square trinomial" is?

  2. anonymous
    • one year ago
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    i think

  3. anonymous
    • one year ago
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    explain it to me just incase

  4. jdoe0001
    • one year ago
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    \(\bf \begin{array}{cccccllllll} {\color{brown}{ a}}^2& + &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ a}}&& &&{\color{blue}{ b}}\\ &\to &({\color{brown}{ a}} + {\color{blue}{ b}})^2&\leftarrow \end{array}\qquad % perfect square trinomial, negative middle term \begin{array}{cccccllllll} {\color{brown}{ a}}^2& - &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ a}}&& &&{\color{blue}{ b}}\\ &\to &({\color{brown}{ a}} - {\color{blue}{ b}})^2&\leftarrow \end{array}\) does it ring a bell?

  5. anonymous
    • one year ago
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    not sure it rings :D

  6. anonymous
    • one year ago
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    well its the one where you identify A B and C right?

  7. jdoe0001
    • one year ago
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    well https://s3.amazonaws.com/ck12bg.ck12.org/curriculum/108348/thumb_540_50.jpg <--- that show it more or less is a trinomial, so it has 3 terms the terms on the extremes, left and right are squared the term in the middle is the multiplication of 2 times the other two terms, without the exponent

  8. anonymous
    • one year ago
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    aaaahhhh i dont know it

  9. jdoe0001
    • one year ago
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    \(\begin{array}{cccccllllll} {\color{brown}{ a}}^2& + &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ a}}&& &&{\color{blue}{ b}}\\ &\to &({\color{brown}{ a}} + {\color{blue}{ b}})^2&\leftarrow \end{array}\qquad % perfect square trinomial, negative middle term \begin{array}{cccccllllll} {\color{brown}{ a}}^2& - &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\ \downarrow && &&\downarrow \\ {\color{brown}{ a}}&& &&{\color{blue}{ b}}\\ &\to &({\color{brown}{ a}} - {\color{blue}{ b}})^2&\leftarrow \end{array}\) notice, "a" and "b" are squared, on the extreme sides the middle is 2 * both, without the exponent

  10. anonymous
    • one year ago
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    so you just combined them but it still means the same thing right

  11. jdoe0001
    • one year ago
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    yes, you'd combine them to a binomial raised at 2 and yes, if you expand the binomial, you'd get back the trinomial

  12. anonymous
    • one year ago
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    so how is this related to the question

  13. jdoe0001
    • one year ago
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    well.. lemme group it a bit

  14. anonymous
    • one year ago
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    okay :)

  15. jdoe0001
    • one year ago
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    \(\bf x^2-2x+y^2-6y=26\implies (x^2-2x)+(y^2-6y)=26 \\ \quad \\ (x^2-2x+{\color{red}{ \square }}^2)+(y^2-6y+{\color{red}{ \square }}^2)=26\) so... we grouped it first and then, we'd want to get a "perfect square trinomial" for those groups any ideas what the missing fellow is there in each?

  16. anonymous
    • one year ago
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    nop :(

  17. jdoe0001
    • one year ago
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    well, think about it recall, the middle term, given here is 2 * both of the other terms

  18. jdoe0001
    • one year ago
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    so the missing number, is really hidden there, in the middle term now if you just factor out the middle term, the missing one will show up

  19. anonymous
    • one year ago
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    is the factor for the first one -1,2

  20. jdoe0001
    • one year ago
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    well, the last number in a perfect trinomial is positive, so... can't be -1 for one

  21. anonymous
    • one year ago
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    yes but the coordinate which gives you the first trionomial

  22. jdoe0001
    • one year ago
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    but let us use say +1 instead the middel term is 2x 2 * x * 1 = 2x so the other terms are "x" and "1" and 2 * both is "2x" so, yes, one is 1

  23. jdoe0001
    • one year ago
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    \(\bf (x^2-2x+{\color{red}{ 1 }}^2)+(y^2-6y+{\color{red}{ \square }}^2)=26\) how about the one for the "y" group?

  24. anonymous
    • one year ago
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    is it 1 as well?

  25. jdoe0001
    • one year ago
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    well, let's see there the middle term in the "y" group is 6y 2 * y * 1 \(\ne 6y\) so, no dice, 2 * y is, 2y, but 2y * 1, is not 6y so is not 1

  26. anonymous
    • one year ago
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    is it 3

  27. jdoe0001
    • one year ago
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    2 * y * 3 = 6y <-- yeap, is 3 now keep in mind that all we're doing is borrowing from our very good fellow Mr Zero, 0 so if we ADD \(1^2\ and \ 3^2\) we also have to SUBTRACT them as well so we end up with \(\bf (x^2-2x+{\color{red}{ 1 }}^2)+(y^2-6y+{\color{red}{ 3 }}^2)-{\color{red}{ 1 }}^2-{\color{red}{ 3 }}^2=26\)

  28. anonymous
    • one year ago
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    okay

  29. jdoe0001
    • one year ago
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    now let us simplify that keeping in mind that, the groups, are not perfect square trinomials thus \(\bf (x^2-2x+{\color{red}{ 1 }}^2)+(y^2-6y+{\color{red}{ 3 }}^2)-{\color{red}{ 1 }}^2-{\color{red}{ 3 }}^2=26 \\ \quad \\ (x^2-2x+1^2)+(y^2-6y+3^2)-1-9=26 \\ \quad \\ (x^2-2x+1^2)+(y^2-6y+3^2)-10=26 \\ \quad \\ (x^2-2x+1^2)+(y^2-6y+3^2)=26+10 \\ \quad \\ (x^2-2x+1^2)+(y^2-6y+3^2)=36 \\ \quad \\ (x-1)^2+(y-3)^2=36\implies (x-1)^2+(y-3)^2=6^2\)

  30. jdoe0001
    • one year ago
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    a circle equation looks like \(\bf (x-{\color{brown}{ h}})^2+(y-{\color{blue}{ k}})^2={\color{purple}{ r}}^2 \qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad radius={\color{purple}{ r}}\) now, can you see the center coordinates and the radius in yours?

  31. anonymous
    • one year ago
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    your explanation is perfect but i dont seem to get it

  32. anonymous
    • one year ago
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    i saw what you did there

  33. jdoe0001
    • one year ago
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    \(\bf (x-{\color{brown}{ 1}})^2+(y-{\color{blue}{ 3}})^2={\color{purple}{ 6}}^2\) see them now?

  34. anonymous
    • one year ago
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    but i dont see the center coordinates

  35. anonymous
    • one year ago
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    oooooohhhhhhh

  36. anonymous
    • one year ago
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    okay :)

  37. jdoe0001
    • one year ago
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    so the center of the circle is at (1,3) and the radius is 6

  38. anonymous
    • one year ago
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    hhhhmmmm thank you so much.you really broke it down for me

  39. jdoe0001
    • one year ago
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    yw

  40. anonymous
    • one year ago
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    i appriciate your time and good explanation

  41. jdoe0001
    • one year ago
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    np

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