Polar Equation to Cartesian Coordinates

[JProgrammer]

I am trying to convert this polar equation to Cartesian coordinates.

r = 8 cos theta

I type the equation into wolfram alpha and it gives me a graph, but no Cartesian points.
If somebody could help me find the cartesian points, I would appreciate it.

Thank you.

 

[MarkFL]

Multiply through by $r$ so that you have:

\(\displaystyle r^2=8r\cos(\theta)\)

Now you should find the conversion to Cartesian coordinates a bit easier. :D

 

[soroban]

Hello, JProgrammer!

If your first idea is to try Wolfram Alpha, I would guess that
ou have never tried one of these conversions.

Here are the basic conversion formulas:

. . [tex]\begin{array}{cc} x \;=\;r\cos\theta \\ y \:=\: r\sin\theta \end{array} \qquad r \:=\:\sqrt{x^2+y^2} [/tex]If the polar equation has a 'naked' [tex]\sin\theta[/tex] or [tex]\cos\theta[/tex]
multiply the equation by [tex]r.[/tex]

If we are given: [tex]r \:=\:6\cos\theta[/tex]

Multiply by [tex]r:\;\;r^2\:=\:6r\cos\theta[/tex]

[tex]\text{Convert: }\;\underbrace{r^2}_{\text{This is }x^2+y^2} \;=\;6\underbrace{r\cos\theta}_{\text{This is }x} [/tex]

[tex]\begin{array}{ccc}\text{So we have:} & x^2+y^2 \:=\:6x \\
& x^2 - 6x + y^2 \:=\:0 \\

& x^2 - 6x + \color{red}{9} \color{black}{\,+\, y^2} \:=\:\color{red}{9} \\

& (x-3)^2 +y^2 \:=\:9 \end{array}[/tex]

We have a circle with center [tex](3,0)[/tex] amd radius [tex]3.[/tex]

 

[JProgrammer]

[tex]\begin{array}{ccc}\text{So we have:} & x^2+y^2 \:=\:6x \\
& x^2 - 6x + y^2 \:=\:0 \\

& x^2 - 6x + \color{red}{9} \color{black}{\,+\, y^2} \:=\:\color{red}{9} \\

& (x-3)^2 +y^2 \:=\:9 \end{array}[/tex]

We have a circle with center [tex](3,0)[/tex] amd radius [tex]3.[/tex]

Okay this all seems to make sense, but where does the 9 come from?

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Okay this all seems to make sense, but where does the 9 come from?

Ignore this, I understand now.

 

[HallsofIvy]

For anyone else who might have wondered, the "9" comes from completing the square. The "perfect square" [tex](x+ a)^2= x^2+ 2a+ a^2[/tex]. Comparing that to [tex]x^2- 6x[/tex] we see that the first two parts will be the same if a= -3. Then [tex]a^2= (-3)^2= 9[/tex]. We can make [tex]x^2- 6x[/tex] a "perfect square" by adding 9: [tex]x^2- 6x+ 9= (x- 3)^2[/tex]. Of course, if we add 9 we must also subtract 9 so we have not changed the value of the expression. [tex]x^2- 6x= x^2- 6x+ 9- 9= (x- 3)^2- 9[/tex].