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hoopshiree

  • one year ago

what is the value of b x^2+y^2+ax+by+c=0

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  1. helder_edwin
    • one year ago
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    the value of \(b\) for what?

  2. hoopshiree
    • one year ago
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    for that equation, \[x ^{2}+y ^{2^{}}+ax+by+c=0\]

  3. hoopshiree
    • one year ago
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    @helder_edwin

  4. helder_edwin
    • one year ago
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    r u sure the question is exactly that? it kind of doesn't make sense

  5. hoopshiree
    • one year ago
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    this graph is part of the question

  6. helder_edwin
    • one year ago
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    u also need to find the values of a, b, and c so that the equation describes the circle of the figure. complete both squares. u know the center and the radius of the circle (see the graph)

  7. hoopshiree
    • one year ago
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    that doesnt help me

  8. hoopshiree
    • one year ago
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    but thanks

  9. helder_edwin
    • one year ago
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    do u know how to complete the square?

  10. hoopshiree
    • one year ago
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    no

  11. helder_edwin
    • one year ago
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    ok i will do it. first the center is C(-2,1) and r=3, do u agree?

  12. hoopshiree
    • one year ago
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    yes

  13. hoopshiree
    • one year ago
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    I got that far.

  14. helder_edwin
    • one year ago
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    ok know, the equation of a circle should be \[\large (x-h)^2+(y-k)^2=r^2 \] we have to make the equation \[\large x^2+y^2ax+by+c=0 \] to resemble this. OK?

  15. helder_edwin
    • one year ago
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    \[\large x^2+y^2+ax+by+c=0 \]

  16. hoopshiree
    • one year ago
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    (x+2)^2+(y-1)^2=3^2

  17. helder_edwin
    • one year ago
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    we know that h=-2, k=1, and r=3, agree?

  18. helder_edwin
    • one year ago
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    yes. expand that!

  19. helder_edwin
    • one year ago
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    can u do it?

  20. hoopshiree
    • one year ago
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    (x+2)+(x+2)+(y-1)+(y-1)=9?

  21. hoopshiree
    • one year ago
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    I'm so confuseddd.

  22. helder_edwin
    • one year ago
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    no \[\large (x+2)^2+(y-1)^2=3^2 \] \[\large (x^2+4x+4)+(y^2-2x+1)=9 \] \[\large x^2+y^2+4x-2y+5=9 \] \[\large x^2+y^2+4x-2y-4=0 \]

  23. helder_edwin
    • one year ago
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    got it?

  24. hoopshiree
    • one year ago
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    so b is 4?

  25. helder_edwin
    • one year ago
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    no, b is the coefficient of \(y\)

  26. hoopshiree
    • one year ago
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    oh -2?

  27. helder_edwin
    • one year ago
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    yes

  28. hoopshiree
    • one year ago
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    can you help me with this one?

  29. helder_edwin
    • one year ago
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    same question?

  30. hoopshiree
    • one year ago
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    have to find a now. so the center is (-1,3) and radius is 4 right?

  31. hoopshiree
    • one year ago
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    (x+1)+(y-3)=16?

  32. RadEn
    • one year ago
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    correction number one, the centre of circle is (1,-2) not (-2,1). so, h = 1 and k = -2 now remember the equation of circle is x^2 + y^2 +ax + by + c = 0 or (x - h)^2 + (y -k)^2 = 0 h,k is the centre of circle. the relation of b and k is b = -1/2(k) so, b = -1/2(-2) = 1

  33. helder_edwin
    • one year ago
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    YES. I am so sorry, my mistake.

  34. hoopshiree
    • one year ago
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    oh boy. okay.

  35. helder_edwin
    • one year ago
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    in the second case the center is (3,-1) and r=4

  36. RadEn
    • one year ago
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    correction again, im not a boy, but a father of that boy ;p

  37. hoopshiree
    • one year ago
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    does A have a certain formula that can be used to find it?

  38. helder_edwin
    • one year ago
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    dont learn formulas, learn concepts. do what i did before: expand \[\large (x-3)^2+(y+1)^2=4^2 \] it should take a minute

  39. SnuggieLad
    • one year ago
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    Welcome To Openstudy

  40. RadEn
    • one year ago
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    a = -1/2 * h

  41. hoopshiree
    • one year ago
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    so (x^2-6x+9)+(y^2+2x+1)=16?

  42. hoopshiree
    • one year ago
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    so a would be = -3 1/2

  43. helder_edwin
    • one year ago
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    a=-6, that is what u got.

  44. ranga
    • one year ago
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    It should be (x^2-6x+9)+(y^2+2y+1)=16 (you had 2x, it should be 2y). In bx^2+y^2+ax+by+c=0, a is the coefficient of x. The coefficient of x in the equation you got is -6 So a = -6.

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