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Math Help - Polar Coordinates-writing an equation for points on a plane

  1. #1
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    Polar Coordinates-writing an equation for points on a plane

    I need to write an equation in polar coordinates for the points in the plane that satisfy
    x^2 - 5x + y^2 - y - 5 = 0
    I have no idea where to begin. Please help.
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  2. #2
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    Does this look right to anyone?

    rcosų^2 - 5(cosų) + ysinų^2 - rsinų - 5=0
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  3. #3
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    Quote Originally Posted by kellylanemccann View Post
    I need to write an equation in polar coordinates for the points in the plane that satisfy
    x^2 - 5x + y^2 - y - 5 = 0
    I have no idea where to begin. Please help.
    x^2 - 5x + y^2 - y - 5 = 0

    x^2 - 5x + \left(-\frac{5}{2}\right)^2 - \left(-\frac{5}{2}\right)^2 + y^2 - y + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 - 5 = 0

    \left(x - \frac{5}{2}\right)^2 - \frac{25}{4} + \left(y - \frac{1}{2}\right) - \frac{1}{4} - 5 = 0

    \left(x - \frac{5}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 - \frac{46}{4} = 0

    \left(x - \frac{5}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{46}{4}

    \left(x - \frac{5}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{\sqrt{46}}{2}\right)^2.


    Now let x = \frac{\sqrt{46}\cos{\theta}}{2} and y = \frac{\sqrt{46}\sin{\theta}}{2}.
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  4. #4
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    Hello, kellylanemccann!

    Convert to polar coordinates: . <br />
x^2 - 5x + y^2 - y - 5 \:=\: 0

    I have no idea where to begin. . . really?
    You are expected to know these conversions: . \begin{array}{c}x \:=\:r\cos\theta \\ y \:=\:r\sin\theta \\ \\[-4mm]x^2+y^2 \:=\:r^2 \end{array}

    \text{We have: }\;\underbrace{x^2 + y^2} \:=\:5x\;\;\;+\;\;\;y\;\; +\;\; 5
    \text{Substitute: }\;\;r^2 \;=\;\overbrace{5r\cos\theta} + \overbrace{r\sin\theta} \;+\; 5

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