# Thread: Polar Coordinates-writing an equation for points on a plane

1. ## 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

2. Does this look right to anyone?

rcosų^2 - 5(cosų) + ysinų^2 - rsinų - 5=0

3. Originally Posted by kellylanemccann
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
$\displaystyle x^2 - 5x + y^2 - y - 5 = 0$

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

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

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

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

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

Now let $\displaystyle x = \frac{\sqrt{46}\cos{\theta}}{2}$ and $\displaystyle y = \frac{\sqrt{46}\sin{\theta}}{2}$.

4. Hello, kellylanemccann!

Convert to polar coordinates: .$\displaystyle x^2 - 5x + y^2 - y - 5 \:=\: 0$

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

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