Answer:
[tex]r(r-4\cos \theta)=0[/tex]
Step-by-step explanation:
We are given the following equation:
[tex]x^2 + y^2 -4x = 0[/tex]
We have to convert it into polar form.
We put
[tex]x = r \cos \theta\\y = r\sin \theta[/tex]
Putting values, we get:
[tex]x^2 + y^2 -4x = 0\\(r\cos \theta)^2 + (r\sin \theta)^2 - 4(r\cos \theta) = 0\\r^2(\cos^2 \theta + \sin^2 \theta) - 4r\cos \theta = 0\\r^2 - 4r\cos \theta = 0\\r(r-4\cos \theta)=0[/tex]
is the required polar form.
