Question

How do you write the given equation `(x-3)^2+y^2=9` into polar form?

Answer 1

`r=6cos theta`

Explanation:

Rewrite the given equation as `x^2 -6x +9 +y^2 =9`. Put `x=r cos theta and y= r sin theta`.

This would make `x^2 +y^2 =r^2` and the equation is thus changed to `r^2 -6rcos theta =0` , Or `r(r-6 cos theta)=0`

Since `r !=0` , the required equation would be `r=6 cos theta`

Answer 2

You substitude `x=rcosθ` and `y=rsinθ`

Explanation:

`(x-3)^2+y^2=9 => (rcosθ-3)^2+(rsinθ)^2=9=>`

`r^2cos^2θ-6rcosθ+9+r^2sin^2θ=9 => r^2-6rcosθ=0 =>`

`r(r-6cosθ)=0=>r-6cosθ=0 =>r=6cosθ`

Answer 3

Expand the square.
Use the conversion equations `x^2+y^2=r^2 and x = rcos(theta)`
Express r as a function of `theta`

Explanation:

Given: `(x-3)^2+y^2=9`

Here is the graph of the original equation:

Expand the square:

`x^2-6x+9+y^2=9`

Use the conversion equations `x^2+y^2=r^2 and x = rcos(theta)`:

`r^2-6rcos(theta)+9=9`

Begin the process of expressing `r` as a function of `theta`

Combine like terms:

`r^2-6rcos(theta)=0`

We can safely divide both sides by r, because we are only eliminating the trivial root `r = 0`:

`r-6cos(theta)=0`

Add `6cos(theta)` to both sides and we shall have `r` as a function of `theta`:

`color(blue)(r=6cos(theta)) larr` The Polar equivalent.

Here is the graph in polar coordinates:

Please observe that the graphs are identical.