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Selorm Group Title

It can be shown that the polar curve r=a\sin(\theta)+b\cos(\theta), where ab\ne 0, is a circle. Find the center (in Cartesian coordinates) and radius of this circle in terms of a and b. x-coordinate of center: y-coordinate of center: Radius:

  • 2 years ago
  • 2 years ago

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  1. ivanmlerner Group Title
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    What is ab\ne 0?

    • 2 years ago
  2. Selorm Group Title
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    r=a sin (theta) + b cos (theta), ab is not = 0

    • 2 years ago
  3. ivanmlerner Group Title
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    Do you agree, that in cartesian coordinates the radius is \[\sqrt{(x-x_0)^2+(y-y_0)^2}\]Where (x0, y0) is the center?

    • 2 years ago
  4. Selorm Group Title
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    yea

    • 2 years ago
  5. myininaya Group Title
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    You could write it as a Cartesian equation by using cos(theta)=x/r and sin(theta)=y/r and r^2=x^2+y^2

    • 2 years ago
  6. Selorm Group Title
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    where does a and b come in?

    • 2 years ago
  7. Selorm Group Title
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    i know that multiplying the equation by r is a legitimate step that can get me r^2=a rsin (theta) + b rcos (theta). Is that right?

    • 2 years ago
  8. myininaya Group Title
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    Yeah sure you can do that.

    • 2 years ago
  9. Selorm Group Title
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    this can give r^2= ay + bx right?

    • 2 years ago
  10. myininaya Group Title
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    Ok and r^2=x^2+y^2

    • 2 years ago
  11. Selorm Group Title
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    yes

    • 2 years ago
  12. Selorm Group Title
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    so that will give x^2+y^2=ay+bx

    • 2 years ago
  13. myininaya Group Title
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    ok and write that equation in this form: (x-h)^2+(y-k)^2=r^2 where (h,k) is center and r is radius

    • 2 years ago
  14. Selorm Group Title
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    so meaning (x-b)^2+(y-a)^2

    • 2 years ago
  15. Selorm Group Title
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    ?

    • 2 years ago
  16. myininaya Group Title
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    hmm....I don't you completed the square right

    • 2 years ago
  17. myininaya Group Title
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    I mean you didn't complete the square...

    • 2 years ago
  18. Selorm Group Title
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    come again

    • 2 years ago
  19. myininaya Group Title
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    \[x^2-bx+y^2-ay=0\] you must complete the square honestly you made some rules up or something

    • 2 years ago
  20. Selorm Group Title
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    ok

    • 2 years ago
  21. Selorm Group Title
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    can u help with that?

    • 2 years ago
  22. myininaya Group Title
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    \[x^2+cx+(\frac{c}{2})^2=(x+\frac{c}{2})^2\]

    • 2 years ago
  23. Selorm Group Title
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    why (c/2)^2 and not (cx/2)^2?

    • 2 years ago
  24. myininaya Group Title
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    \[(x+\frac{c}{2})^2=(x+\frac{c}{2})(x+\frac{c}{2})=x(x+\frac{c}{2})+\frac{c}{2}(x+\frac{c}{2})\] \[=x^2+\frac{c}{2}x+\frac{c}{2}x+(\frac{c}{2})^2=x^2+cx+(\frac{c}{2})^2\]

    • 2 years ago
  25. Selorm Group Title
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    ok

    • 2 years ago
  26. Selorm Group Title
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    what next?

    • 2 years ago
  27. myininaya Group Title
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    \[x^2-bx+y^2-ay=0 \]

    • 2 years ago
  28. myininaya Group Title
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    You need to complete the square for both the x part and the y part.

    • 2 years ago
  29. Selorm Group Title
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    so we get (x+c/2)^2+(y+b/2)^2=0?

    • 2 years ago
  30. Selorm Group Title
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    then what next? equate each part to 0?

    • 2 years ago
  31. myininaya Group Title
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    hmmm.... So you have x^2-bx+b/2=(x-b/2)^2

    • 2 years ago
  32. myininaya Group Title
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    But whatever you add to one side you add to the other .

    • 2 years ago
  33. Selorm Group Title
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    x=b/2,y =a/2 and r=sqrt(a^2+b^2)/2

    • 2 years ago
  34. myininaya Group Title
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    YEP that is right.

    • 2 years ago
  35. Selorm Group Title
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    thank you

    • 2 years ago
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