Question

How do you convert ${x^2} + {\left( {y - 4} \right)^2} = 16$ to polar form?

Answer 1

Hint: In order to obtain the polar form of the given equation , substitute the variable $x$ by $r\cos \theta $and the variable $y$by $r\sin \theta $. Now simplify the equation using trigonometric identity as ${\cos ^2}\theta + {\sin ^2}\theta = 1$and combine the like terms to obtain the required result.

Complete step by step answer:
We are given a quadratic equation in two variables i.e. $x\,and\,y$ as
${x^2} + {\left( {y - 4} \right)^2} = 16$
There are two ways to determine an equation , one is cartesian form and the other is polar form .
In our question we are given the equation in cartesian form and we have to obtain its polar form .
So in order to convert any cartesian plane equation into its equivalent polar form we have to substitute variable $x = r\cos \theta $ and $y = r\sin \theta $ in the cartesian plane equation.
Now putting $x = r\cos \theta $and $y = r\sin \theta $in our given equation , we get
${\left( {r\cos \theta } \right)^2} + {\left( {r\sin \theta - 4} \right)^2} = 16$
Using the formula of ${\left( {A - B} \right)^2} = {A^2} + {B^2} - 2AB$ to expand the second term
${r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta + 16 - 8r\sin \theta = 16$
Pulling out common from the first two terms and subtracting the number $16$ from both sides of the equation, we obtain our equation as
${r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) + 16 - 8r\sin \theta - 16 = 16 - 16$
Now using the identity of trigonometry as ${\cos ^2}\theta + {\sin ^2}\theta = 1$
$
  {r^2}\left( 1 \right) - 8r\sin \theta = 0 \\
  {r^2} - 8r\sin \theta = 0 \\
 $
Simplifying further ,we have
${r^2} = 8r\sin \theta $
Now dividing both sides of the equation with $r$, we get
$
  \dfrac{{{r^2}}}{r} = \dfrac{{8r\sin \theta }}{r} \\
  r = 8\sin \theta \\
 $

Therefore, the polar form of the given equation is equal to $r = 8\sin \theta $.

Note: Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
The graph of polar form and cartesian form of every equation is always the same.
Make sure to simplify the end result.
The equation given in our question is an equation of a circle.