# Thread: converting polar equation to cartesian equation

1. ## converting polar equation to cartesian equation

With this problem it ask to convert r=-3 cos(theta) from polar form to cartesian

This is what I get
$\displaystyle \begin{array}{l} r = - 3\sin (\theta ) \\ (x - \frac{3}{2})(x - \frac{3}{2}) + (\frac{{ - 3}}{2})^2 + y^2 = 0 + (\frac{{ - 3}}{2})^2 \\ (x - \frac{3}{2})^2 + (\frac{{ - 3}}{2})^2 + y^2 = (\frac{{ - 3}}{2})^2 \\ \end{array}$

However the answer given is as below. What have I done wrong?
$\displaystyle (x - \frac{3}{2})^2 + y^2 = (\frac{{ - 3}}{2})^2 \\$

2. Hello,

If the polar coordinate is defined as:
x=r cos(theta)
y=r sin(theta)

then,
r=\sqrt{x^2+y^2},
cos(theta)=x/r,
sin(theta)=y/r.

By substituton,
r=-3 cos(theta) becomes
r=-3x/r that is, r^2=-3x
so that x^2+y^2=-3x.
((x+3/2)^2+y^2=(3/2)^2.)

Please adjust the above argument to the appropriate polar coordinate you are using...

Bye.

4. Hello,

Originally Posted by Craka
Sorry I couldn't figure out what you had written.
Would you explain how you got these?
(If you are asked to convert r=-3 cos(theta), why begin with r=-3sin(theta)? Where does the second line come from?)

Besides, there are several ways to define coordinates.
What is the definition (in your book) of the polar coordinates?

Bye.

5. Hello, Craka!

Convert $\displaystyle r\:=\:-3\cos\theta$ to cartesian form.

$\displaystyle \text{Multiply by }r\!:\;\;\underbrace{r^2} \;\;=\;-3\underbrace{r\cos\theta}$

. . . . . . . . .$\displaystyle x^2+y^2 \;=\quad\; -3x$

Then: .$\displaystyle x^2 + 3x + y^2 \;=\;0$

. . $\displaystyle x^2 + 3x + {\color{blue}\frac{9}{4}} + y^2 \;=\;{\color{blue}\frac{9}{4}}$

. . . .$\displaystyle \boxed{\left(x + \frac{3}{2}\right)^2 + y^2 \;=\;\frac{9}{4}}$

This is a circle with center $\displaystyle \left(-\frac{3}{2},\:0\right)$ and radius $\displaystyle \frac{3}{2}$