Proving that $\rho =3\cos\varphi$ is a circle Practising some exercises in Physics. In one of the questions they said that $\rho =3\cos\varphi$ is a circle and asked to prove that it is a circle.
I'm trying to figure out how to do it. On the internet I saw that the general form of a circle in central $r_0,\varphi$ and radius $a$ is:
$$
r^2 - 2 r r_0 \cos(\theta - \varphi) + r_0^2 = a^2
$$
Do I need to compare the two?

The relation between rectangular and polar coordintes are:
$x=\rho \cos \phi$
$y=\rho \sin \phi$
$\rho=\sqrt{x^2+y^2}$
We have:
$\rho=2a\cos\phi$
So:
$\sqrt{x^2+y^2}=2a\frac{x}{\sqrt{x^2+y^2}}$
$\Rightarrow x^2+y^2-2ax=0$
Or:
$x^2-2ax+a^2+y^2=a^2$
which gives:
$(x-a)^2+y^2=a^2$
which is the equation of a circle center at (a, 0). In your question $2a=3$ so the equation of circle is:
$(x-\frac32)^2+y^2=\frac 94$
