MHB 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:

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

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:

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

If we are given: r \:=\:6\cos\theta

Multiply by r:\;\;r^2\:=\:6r\cos\theta

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

\begin{array}{ccc}\text{So we have:} &amp; x^2+y^2 \:=\:6x \\<br /> &amp; x^2 - 6x + y^2 \:=\:0 \\<br /> <br /> &amp; x^2 - 6x + \color{red}{9} \color{black}{\,+\, y^2} \:=\:\color{red}{9} \\<br /> <br /> &amp; (x-3)^2 +y^2 \:=\:9 \end{array}

We have a circle with center (3,0) amd radius 3.

 

[JProgrammer]

\begin{array}{ccc}\text{So we have:} &amp; x^2+y^2 \:=\:6x \\<br /> &amp; x^2 - 6x + y^2 \:=\:0 \\<br /> <br /> &amp; x^2 - 6x + \color{red}{9} \color{black}{\,+\, y^2} \:=\:\color{red}{9} \\<br /> <br /> &amp; (x-3)^2 +y^2 \:=\:9 \end{array}

We have a circle with center (3,0) amd radius 3.

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

- - - Updated - - -

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" (x+ a)^2= x^2+ 2a+ a^2. Comparing that to x^2- 6x we see that the first two parts will be the same if a= -3. Then a^2= (-3)^2= 9. We can make x^2- 6x a "perfect square" by adding 9: x^2- 6x+ 9= (x- 3)^2. Of course, if we add 9 we must also subtract 9 so we have not changed the value of the expression. x^2- 6x= x^2- 6x+ 9- 9= (x- 3)^2- 9.