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

• Feb 15th 2010, 09:56 PM
kellylanemccann
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
• Feb 15th 2010, 10:58 PM
kellylanemccann
Does this look right to anyone?

rcosų^2 - 5(cosų) + ysinų^2 - rsinų - 5=0
• Feb 15th 2010, 11:28 PM
Prove It
Quote:

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

$x^2 - 5x + y^2 - y - 5 = 0$

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

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

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

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

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

Now let $x = \frac{\sqrt{46}\cos{\theta}}{2}$ and $y = \frac{\sqrt{46}\sin{\theta}}{2}$.
• Feb 16th 2010, 05:38 AM
Soroban
Hello, kellylanemccann!

Quote:

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

I have no idea where to begin. . . really?

You are expected to know these conversions: . $\begin{array}{c}x \:=\:r\cos\theta \\ y \:=\:r\sin\theta \\ \\[-4mm]x^2+y^2 \:=\:r^2 \end{array}$

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