Question

what is the polar form of the equation?$(x + 2) ^{2} + y ^{2} = 4$

Answer 1

ANSWER

$r(r+4\cos \theta)=0$

EXPLANATION

We want to convert the given equation to polar form:

$(x+2)^2+y^2=4$

The first step is to expand the bracket:

$\begin{gathered} (x+2)(x+2)+y^2=4 \\ x^2+2x+2x+4+y^2=4 \\ x^2+4x+y^2=4-4 \\ x^2+4x+y^2=0 \end{gathered}$

Now, we will convert it by carrying out the following transformations:

$\begin{gathered} x=r\cos \theta \\ y=r\sin \theta \\ x^2+y^2=r^2 \end{gathered}$

The equation then becomes:

$\begin{gathered} x^2+4(r\cos \theta)+y^2=0 \\ \Rightarrow x^2+y^2+4r\cos \theta=0 \end{gathered}$

Therefore, it becomes:

$\begin{gathered} r^2+4\cos \theta=0 \\ \Rightarrow r(r+4\cos \theta)=0 \end{gathered}$

That is the polar form of the equation.