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:

. . $$\begin{array}{cc} x \;=\;r\cos\theta \\ y \:=\: r\sin\theta \end{array} \qquad r \:=\:\sqrt{x^2+y^2}$$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 $$r:\;\;r^2\:=\:6r\cos\theta$$

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

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

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

 

[JProgrammer]

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

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?

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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" $$(x+ a)^2= x^2+ 2a+ a^2$$. 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. $$x^2- 6x= x^2- 6x+ 9- 9= (x- 3)^2- 9$$.