How do you convert ${x^2} + {y^2} + 2x = 0$ to polar form?

Hint: According to the question we have to convert the given quadratic expression ${x^2} + {y^2} + 2x = 0$ which is as mentioned in the question in the polar form. So, first of all to convert the given quadratic expression in the question in the polar form we have to understand about the polar form which is as explained below:
Polar form: Polar form of a complex number is a different way to represent a complex number apart from a rectangular form and we represent the complex number in the form of $z = a + ib$ where i is the imaginary number but in the polar form, complex numbers are represented as the combination of arguments.
Now, to determine the required polar form of the given expression we have to substitute the variables which are x and y according to the given expression with the complex form.
Now, we have to solve the expression as obtained after substituting all the complex values in the expression which is ${x^2} + {y^2} + 2x = 0$.
Now, to solve the polar expression in more simpler form we have to use the identity which is as mentioned below:
Identity used:
$ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1................(A)$
Hence, with the help of the identity above we can easily determine the required polar form of the given expression.

Complete step-by-step solution:
Step 1: First of all to convert the given quadratic expression in the question into the polar form we have to understand about the polar form which is as explained in the solution hint.
Step 2: Now, to determine the required polar form of the given expression we have to substitute the variables which are x and y according to the given expression with the complex form. Hence,
\[
   \Rightarrow x = r\cos \theta \\
   \Rightarrow y = r\sin \theta \\
 \]
Step 3: Now, we have to sole the expression as obtained after substituting all the complex values in the expression which is${x^2} + {y^2} + 2x = 0$. Hence,
$ \Rightarrow {(r\cos \theta )^2} + {(r\sin \theta )^2} - 2r\cos \theta = 0$
Step 4: Now, to solve the polar expression in more simpler form we have to use the identity (A) which is as mentioned in the solution hint. Hence,
$
   \Rightarrow {r^2}[{(\cos \theta )^2} + {(\sin \theta )^2}] - 2r\cos \theta = 0 \\
   \Rightarrow {r^2} - 2r\cos \theta = 0 \\
 $
Now, we just have to rearrange the terms of the polar expression as obtained just above,
$ \Rightarrow 2r\cos \theta = {r^2}$

Hence, with the help of the identity we have determined the required polar form of the expression is $ \Rightarrow 2r\cos \theta = {r^2}$.

Note: Polar form of a complex number is a different way to represent a complex number apart from a rectangular form and we represent the complex number in the form of $z = a + ib$ where I is the imaginary number but in the polar form.
To make the polar expression more simpler we have to use the identity which is ${\sin ^2}\theta + {\cos ^2}\theta = 1$ which is as mentioned in the solution hint.