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

How do you convert x2+(y−4)2=16{x^2} + {\left( {y - 4} \right)^2} = 16 to polar form?

Explanation

Solution

In order to obtain the polar form of the given equation , substitute the variable xx by rcos⁡θr\cos \theta and the variable yyby rsin⁡θr\sin \theta . Now simplify the equation using trigonometric identity as cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1and 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 yx\,and\,y as
x2+(y−4)2=16{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=rcos⁡θx = r\cos \theta and y=rsin⁡θy = r\sin \theta in the cartesian plane equation.
Now putting x=rcos⁡θx = r\cos \theta and y=rsin⁡θy = r\sin \theta in our given equation , we get
(rcos⁡θ)2+(rsin⁡θ−4)2=16{\left( {r\cos \theta } \right)^2} + {\left( {r\sin \theta - 4} \right)^2} = 16
Using the formula of (A−B)2=A2+B2−2AB{\left( {A - B} \right)^2} = {A^2} + {B^2} - 2AB to expand the second term
r2cos⁡2θ+r2sin⁡2θ+16−8rsin⁡θ=16{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 1616 from both sides of the equation, we obtain our equation as
r2(cos⁡2θ+sin⁡2θ)+16−8rsin⁡θ−16=16−16{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⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1
r2(1)−8rsin⁡θ=0 r2−8rsin⁡θ=0  {r^2}\left( 1 \right) - 8r\sin \theta = 0 \\\ {r^2} - 8r\sin \theta = 0 \\\
Simplifying further ,we have
r2=8rsin⁡θ{r^2} = 8r\sin \theta
Now dividing both sides of the equation with rr, we get
r2r=8rsin⁡θr r=8sin⁡θ  \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=8sin⁡θr = 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.