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?
 A: The best thing to do is to swap from polar coordinates $(\rho,\phi)$ to Cartesian coordinates $(x,y)$.
Recall that $\rho=\sqrt{x^2+y^2}$ (remember $r\geq0$!), so $\rho=\sqrt{x^2+y^2}=3\cos\phi$.
As for dealing with the trigonometric term, I find it best to consider the case when $\phi$ is acute and visualise the right-angled triangle it forms in the positive quadrant. Then, $\cos\phi=\frac{x}{\sqrt{x^2+y^2}}$, so the equation further becomes $x^2+y^2=3x$.
Lastly, we can complete the square to get $(x^2-3x+\frac94)+y^2=(x-\frac32)^2+y^2=(\frac32)^2$, indicating that this is a circle with centre $\boxed{\left(\frac32,0\right)}$ and radius $\boxed{\frac32}$.
A: You are given
$\rho = 3 \cos \varphi $
From which it follows that
$x = \rho \cos \varphi = 3 \cos^2 \varphi $
and
$y = \rho \sin \varphi = 3 \cos \varphi \sin \varphi $
Simplifying the expressions, we get
$x = \frac{3}{2} (1 + \cos 2 \varphi) $
and
$y = \frac{3}{2} \sin 2 \varphi $
Thus
$\cos 2 \varphi = -1 + \dfrac{2}{3} x $
and
$\sin 2 \varphi = \frac{2}{3} y $
And since $\cos^2 \theta + \sin^2 \theta = 1$ for any $\theta$ , then
$\left(-1 + \dfrac{2}{3} x \right)^2 + \left(\dfrac{2}{3} y \right)^2 = 1$
Multiplying through by $\left( \dfrac{3}{2} \right)^2$ , we get,
$ \left(x - \dfrac{3}{2} \right)^2 + y^2 = \left( \dfrac{3}{2} \right)^2 $
Which is an equation of a circle centered at $ \left(\dfrac{3}{2}, 0 \right) $ and having a radius of $\left( \dfrac{3}{2} \right) $.
A: 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:
$(x-\frac32)^2+y^2=\frac 94$
