Question

How do you convert \[{x^2} + {y^2} - 2x = 0\] to polar form?

Answer 1

Hint: We will use the polar equation form to find the required equation for the above Cartesian equation.
When we think about plotting points in the plane, we usually think of rectangular coordinates \[(x,y)\] in the Cartesian coordinate plane.
However, there are other ways of writing a coordinate pair and other types of grid system.
Polar coordinates are points labelled \[(r,\theta )\]and plotted on a polar grid.
The polar grid is represented as a series of concentric circles radiating out from the pole, or the origin of the coordinate plane.

Formula used:
To convert from polar to rectangular (Cartesian) coordinates use the following formulas (derived from their trigonometric function definitions):
So, we can write that: \[\cos \theta = \dfrac{x}{r}\].
Using cross multiplication, we get:
\[ \Rightarrow x = r\cos \theta ..............(1)\].
Again, we can write that:
\[ \Rightarrow \sin \theta = \dfrac{y}{r}\].
Using cross multiplication, we get:
\[ \Rightarrow y = r\sin \theta .............(2)\].
So,
After squaring the equation\[(1)\], we get:
\[ \Rightarrow {x^2} = {r^2}{\cos ^2}\theta ..............(3)\].
After squaring the both sides of the equation\[(2)\], we get:
\[ \Rightarrow {y^2} = {r^2}{\sin ^2}\theta ..............(4)\].
Now, add equation \[(3)\]and \[(4)\], we get:
\[ \Rightarrow {x^2} + {y^2} = {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta .\]
After re-arranging, we get:
\[ \Rightarrow {x^2} + {y^2} = {r^2}({\cos ^2}\theta + {\sin ^2}\theta ).\]
But we know that \[({\cos ^2}\theta + {\sin ^2}\theta ) = 1.\]
So, if we put this into the above equation, we get:
\[ \Rightarrow {x^2} + {y^2} = {r^2} \times 1\].
Or, we can write the following form:
\[ \Rightarrow {x^2} + {y^2} = {r^2}.\]
Another formula:
if we divide equation\[(2)\]by equation\[(1)\], we get:
\[ \Rightarrow \dfrac{{r\sin \theta }}{{r\cos \theta }} = \dfrac{y}{x}\].
Or, we can re-write it as following:
\[ \Rightarrow \tan \theta = \dfrac{y}{x}\].

Complete step by step answer:
The given equation in the question is: \[{x^2} + {y^2} - 2x = 0\]
So, we will put the polar equation of the Cartesian axis’s value in the above equation.
So, we will put \[x = r\cos \theta \] and \[y = r\sin \theta \].
By putting these values, we get:
\[ \Rightarrow {(r\cos \theta )^2} + {(r\sin \theta )^2} - 2.r\cos \theta = 0\].
Now, by doing the squared in the terms, we get:
\[ \Rightarrow {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta - 2r\cos \theta = 0\].
Now, by simplifying the above equation, we get:
\[ \Rightarrow {r^2}({\cos ^2}\theta + {\sin ^2}\theta ) - 2r\cos \theta = 0\].
Now, using the formula\[({\cos ^2}\theta + {\sin ^2}\theta ) = 1\], we get:
\[ \Rightarrow {r^2} \times 1 - 2r\cos \theta = 0\].
Now, multiplying the terms, we get:
\[ \Rightarrow {r^2} - 2r\cos \theta = 0\].
Now, add ‘\[2.r\cos \theta \]’ on the both side of the equation, we get:
\[ \Rightarrow {r^2} - 2r\cos \theta + 2r\cos \theta = 2r\cos \theta \].
Now, simplifying the above equation, we get:
\[ \Rightarrow {r^2} = 2r\cos \theta ............(1)\].
As, \[r \ne 0\], divide both the sides by \[r\], we get:
\[ \Rightarrow \dfrac{{{r^2}}}{r} = \dfrac{{2r\cos \theta }}{r}\].
Now, simplify the above equation, we get:
\[ \Rightarrow r = 2\cos \theta ...........(2)\].
Therefore, equation\[(1)\] and \[(2)\] both are the polar equation form.

\[\therefore \] \[r = 2\cos \theta \] or \[{r^2} = 2.r\cos \theta \] is the required polar form.

Note: Alternative way of solution:
Given equation: \[{x^2} + {y^2} - 2x = 0\].
Now, divide both the sides of the equation by \[{x^2}\], we get:
\[ \Rightarrow \dfrac{{{x^2} + {y^2} - 2x}}{{{x^2}}} = \dfrac{0}{{{x^2}}}\], as \[x \ne 0\].
Now, by simplifying it, we get:
\[ \Rightarrow \dfrac{{{x^2}}}{{{x^2}}} + \dfrac{{{y^2}}}{{{x^2}}} - \dfrac{{2x}}{{{x^2}}} = 0\].
Now, further simplify it, we get:
\[ \Rightarrow 1 + \dfrac{{{y^2}}}{{{x^2}}} - \dfrac{2}{x} = 0.............(1)\].
By the formula we know that: \[\dfrac{y}{x} = \tan \theta \].
Now, put this value in equation\[(1)\], we get:
\[ \Rightarrow 1 + {\left( {\tan \theta } \right)^2} - \dfrac{2}{x} = 0............(2)\].
We know that: \[1 + {\tan ^2}\theta = {\sec ^2}\theta \].
Put this value in equation\[(2)\], we get:
\[ \Rightarrow {\sec ^2}\theta - \dfrac{2}{x} = 0\].
Now, put polar equation\[x = r\cos \theta \], we get:
\[ \Rightarrow {\sec ^2}\theta - \dfrac{2}{{r\cos \theta }} = 0\].
Simplify it, we get:
\[ \Rightarrow {\sec ^2}\theta - \dfrac{2}{r}\sec \theta = 0\].
As, \[\sec \theta \ne 0\], divide the above equation by \[\sec \theta \], we get:
\[ \Rightarrow \sec \theta - \dfrac{2}{r} = 0\].
Now, take the ‘\[\dfrac{2}{r}\]’into the R.H.S, we get:
\[ \Rightarrow \sec \theta = \dfrac{2}{r}\].
Now, by doing cross multiplication, we get:
\[ \Rightarrow r = \dfrac{2}{{\sec \theta }}\].
Now, simplify it further, we get:
\[ \Rightarrow r = 2\cos \theta \], as \[\cos \theta = \dfrac{1}{{\sec \theta }}\].
Therefore, the required polar form is \[r = 2\cos \theta \].