Question

How do you convert `x^2 +y^2-2x=0` to polar form?

Answer 1

`2rcos theta = r^2`

Explanation:

Substitute:

`{ (x= r cos theta), (y=r sin theta) :}`

So we have:

`0 = x^2+y^2-2x`

`=(r cos theta)^2+(r sin theta)^2-2r cos theta`

`=r^2(cos^2 theta + sin^2 theta) - 2r cos theta`

`=r^2-2r cos theta`

Add `2rcos theta` to both ends and transpose to find:

`2rcos theta = r^2`

graph{x^2+y^2-2x=0 [-1.667, 3.333, -1.28, 1.22]}