Question

How do you convert $ {{x}^{2}}+{{y}^{2}}=49 $ into polar form?

Answer 1

Hint: In this question, we have to convert an equation into polar form. As we know, the given equation is in the rectangular form. So, we first change the rectangular form into a circle equation and then convert it into polar form. That is, we will rewrite the rectangular equation into circle equation $ {{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}} $ , where (h, k) is the center of the circle and r is the radius of the circle, and then we will apply the polar form method. We will let $ x=r\cos \theta $ and $ y=r\sin \theta $ in the equation and then make further calculations. Then, we will apply the trigonometric identity $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $ on the LHS of the equation and take the square root to get the solution to the problem.

Complete step by step answer:
According to the question, we have to convert a rectangular into a polar form
Thus, we will apply the polar form method to do the same.
The equation given to us is $ {{x}^{2}}+{{y}^{2}}=49 $ --------- (1)
So, first, we will rewrite the equation (1) into circle equation, that is
 $ {{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}} $ , where (h, k) is the center of the circle and r is the radius of the circle
Therefore, equation (1) will be,
 $ \Rightarrow {{(x-0)}^{2}}+{{(y-0)}^{2}}={{7}^{2}} $
 $ \Rightarrow {{(x)}^{2}}+{{(y)}^{2}}={{7}^{2}} $ --------- (2)
Implies, the center of the circle is (0, 0) and radius is equal to 7.
So, now we will apply the polar form method in equation (2), by letting $ x=r\cos \theta $ and $ y=r\sin \theta $ in equation (2), we get
 $ \Rightarrow {{(r\cos \theta )}^{2}}+{{(r\sin \theta )}^{2}}={{7}^{2}} $
On solving the above equation, we get
 $ \begin{align}
  & \Rightarrow {{(r)}^{2}}{{(\cos \theta )}^{2}}+{{(r)}^{2}}{{(\sin \theta )}^{2}}=49 \\
 & \Rightarrow {{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta =49 \\
\end{align} $
Now, we will take common $ {{r}^{2}} $ on the left-hand side in the above equation, we get
 $ \Rightarrow {{r}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )=49 $
Now, we will apply the trigonometric identity $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $ on the left-hand side in the above equation, we get
 $ \Rightarrow {{r}^{2}}.(1)=49 $
Now, we will take the square root on both sides in the above equation, we get
 $ \Rightarrow \sqrt{{{r}^{2}}}=\sqrt{49} $
On further solving, we get
 $ \Rightarrow r=\pm 7 $ which is the required solution to the problem.
Therefore, for the equation $ {{x}^{2}}+{{y}^{2}}=49 $ , the value of its polar form is $ r=+7 $ and $ r=-7 $.

Note:
 While solving this problem, keep in mind that the RHS on the given equation is equal to $ {{r}^{2}} $ and not equal to $ r $ . Do mention all the identities and formulas to avoid errors in the calculations. The polar form of the equation must be equal to $ \pm $ and not only $ + $ or $ - $ sign.