Question

How do you convert ${{x}^{2}}+{{y}^{2}}=5$ to polar form?

Answer 1

Hint: For finding the polar form of the given equation ${{x}^{2}}+{{y}^{2}}=5$ , we will have to follow the some steps as:
First, we will consider $x$ and $y$ as $r\cos \theta $ and $r\sin \theta $ respectively.
Then, we will replace $x$ and $y$in the given equation as $r\cos \theta $ and $r\sin \theta $ respectively.After that we will substitute $\theta $ from the equation. Then we will form the polar form of the given equation.

Complete step by step answer:
Since, we have the question in form of $x$ and $y$ in the form of $r$ and $\theta $ as:
$\Rightarrow {{x}^{2}}+{{y}^{2}}=5$
Let us consider $x$ and $y$ as:
$\Rightarrow x=r\cos \theta $
And
$\Rightarrow y=r\sin \theta $
Now, we will replace the value of $x$ and $y$ in the form of $r$ and $\theta $ in the given equation ${{x}^{2}}+{{y}^{2}}=5$ as:
$\Rightarrow {{\left( r\cos \theta \right)}^{2}}+{{\left( r\sin \theta \right)}^{2}}=5$
Now, will open the bracket in which we will have to square $r\cos \theta $ and $r\sin \theta $ as:
$\Rightarrow {{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta =5$
Since, ${{r}^{2}}$ is common in both terms ${{r}^{2}}{{\cos }^{2}}\theta $ and ${{r}^{2}}{{\sin }^{2}}\theta $ . So, we can write the above equation as:
$\Rightarrow {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)=5$
Here, we can write the above equation as:
$\Rightarrow {{r}^{2}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)=5$
Since, we know that the formula of trigonometry identities, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ , we can convert the above equation as:
$\Rightarrow {{r}^{2}}\left( 1 \right)=5$
$\Rightarrow {{r}^{2}}\times 1=5$
After multiplication the above equation will be as:
$\Rightarrow {{r}^{2}}=5$
Now, we will find the square root of the above equation and the square root of the $5$ is $\sqrt{5}$.
\[\Rightarrow \sqrt{{{r}^{2}}}=\sqrt{5}\]
$\Rightarrow r=\sqrt{5}$
Hence, the standard for the given equation ${{x}^{2}}+{{y}^{2}}=5$ is $r=\sqrt{5}$ .

Note:
Here we will check whether the solution of the given equation is correct or not in the following way:
Since, we have the standard form of the given question $r=\sqrt{5}$.
After squaring both sides, the equation will be as:
$\Rightarrow {{r}^{2}}=5$
Now, we can write the above equation as:
$\Rightarrow {{r}^{2}}\times 1=5$
Now, we will use the trigonometry identities formula that is ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ . So, the above equation will be:
$\Rightarrow {{r}^{2}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)=5$
Here we will have the above equation as:
$\Rightarrow {{r}^{2}}{{\sin }^{2}}\theta +{{r}^{2}}{{\cos }^{2}}\theta =5$
Now, we consider $r\cos \theta =x$ and $r\sin \theta =y$. We will apply it and organize it to get the given equation as:
$\Rightarrow {{\left( r\cos \theta \right)}^{2}}+{{\left( r\sin \theta \right)}^{2}}=5$
Since, we got the normal equation from the standard equation. Hence, the solution is correct.