Question

How do you write the given equation ${\left( {x - 3} \right)^2} + {y^2} = 9$ into polar form?

Answer 1

Here we will first expand the square of the equation by using the algebraic identity. Then we will put the value of the polar coordinates in the equation and solve it. Then after solving we will get the polar form of the given equation.

Complete Step by Step Solution:
The given equation is ${\left( {x - 3} \right)^2} + {y^2} = 9$.
First, we will expand the square of the equation by simply using the algebraic identity i.e. ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$. Therefore, we get
$\Rightarrow {x^2} + {3^2} - 2 \times x \times 3 + {y^2} = 9$
$\Rightarrow {x^2} + 9 - 6x + {y^2} = 9$
Subtracting 9 from both sides, we get
$\Rightarrow {x^2} + {y^2} - 6x = 0$
Now to convert the given equation into the polar form we will substitute $x = r\cos \theta$ and $y = r\sin \theta$. Therefore, we get
$\Rightarrow {\left( {r\cos \theta } \right)^2} + {\left( {r\sin \theta } \right)^2} - 6\left( {r\cos \theta } \right) = 0$
Applying the exponent on the terms, we get
$\Rightarrow {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta - 6r\cos \theta = 0$
Now we will take ${r^2}$ common from the first two terms of the equation, so we get
$\Rightarrow {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) - 6r\cos \theta = 0$
We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$. Therefore, substituting this value in the above equation, we get
$\Rightarrow {r^2}\left( 1 \right) - 6r\cos \theta = 0$
$\Rightarrow {r^2} - 6r\cos \theta = 0$
Now taking $r$ common from the terms, we get
$\Rightarrow r\left( {r - 6\cos \theta } \right) = 0$
As the value of $r$ can never be zero i.e. $r \ne 0$. So, the required equation, will be
$\Rightarrow r - 6\cos \theta = 0$
$\Rightarrow r = 6\cos \theta$

Hence the given equation ${\left( {x - 3} \right)^2} + {y^2} = 9$ into polar form can be written as $r = 6\cos \theta$.

Note:
Here we have to note that the polar coordinate system is the system in which the coordinates of a point is represented by the distance of that point from a reference point and by the angle from the reference plane. But the normal rectangular coordinates system is the system in which the coordinates of a point is represented by the distance of the point with respect to the X-axis and Y-axis while writing the coordinates of a point always the X-axis intercept of the point is written first and then the Y-axis intercept was written. This axis are the infinite lines in the Cartesian plane.