
How do you write $ r = 4\cos \theta $ into rectangular form?
Answer
443.7k+ views
Hint: To find the rectangular form of the equation from the polar form, here we will use the formula $ x = r\cos \theta $ and $ y = r\sin \theta $ then finding the correlation between the two and will simplify accordingly for the required resultant solution.
Complete step-by-step solution:
Take the given expression: $ r = 4\cos \theta $
Multiply the above equation by “r” on both the sides of the equation.
$ \Rightarrow {r^2} = 4r\cos \theta $ …. (A)
We have following equations in the polar form –
$
x = r\cos \theta \\
y = r\sin \theta \\
$ …(B)
From the above two equations we can derive the third correlation –
$ {x^2} + {y^2} = {r^2} $ …. (C)
From equation (A) and (C)
$ \Rightarrow {x^2} + {y^2} = 4x $
Move the term from the right-hand side of the equation to the left hand side of the equation. When you move any term from one side to another, the sign of the term also changes. Positive terms become negative and vice-versa.
$ \Rightarrow {x^2} + {y^2} - 4x = 0 $
Rearrange the above equation.
$ \Rightarrow {x^2} - 4x + {y^2} = 0 $
Add the number on both the sides of the above equation.
$ \Rightarrow {x^2} - 4x + 4 + {y^2} = 4 $
Make the group of the first three terms in the above equation. And using the identity for $ {(a - b)^2} = {a^2} - 2ab + {b^2} $
$ \Rightarrow {(x - 2)^2} + {y^2} = 4 $
This is the required solution.
Note: Be careful while solving and simplifying the equations. Always remember that when you add, subtract on one side of the equation, you have to add on both the sides of the equation. Equivalent form of the equation should be formed. Always remember the correlation between rectangular and polar form. Always remember when you move any term from one side to another then the sign of the term also changes. Positive term becomes negative and negative term becomes positive.
Complete step-by-step solution:
Take the given expression: $ r = 4\cos \theta $
Multiply the above equation by “r” on both the sides of the equation.
$ \Rightarrow {r^2} = 4r\cos \theta $ …. (A)
We have following equations in the polar form –
$
x = r\cos \theta \\
y = r\sin \theta \\
$ …(B)
From the above two equations we can derive the third correlation –
$ {x^2} + {y^2} = {r^2} $ …. (C)
From equation (A) and (C)
$ \Rightarrow {x^2} + {y^2} = 4x $
Move the term from the right-hand side of the equation to the left hand side of the equation. When you move any term from one side to another, the sign of the term also changes. Positive terms become negative and vice-versa.
$ \Rightarrow {x^2} + {y^2} - 4x = 0 $
Rearrange the above equation.
$ \Rightarrow {x^2} - 4x + {y^2} = 0 $
Add the number on both the sides of the above equation.
$ \Rightarrow {x^2} - 4x + 4 + {y^2} = 4 $
Make the group of the first three terms in the above equation. And using the identity for $ {(a - b)^2} = {a^2} - 2ab + {b^2} $
$ \Rightarrow {(x - 2)^2} + {y^2} = 4 $
This is the required solution.
Note: Be careful while solving and simplifying the equations. Always remember that when you add, subtract on one side of the equation, you have to add on both the sides of the equation. Equivalent form of the equation should be formed. Always remember the correlation between rectangular and polar form. Always remember when you move any term from one side to another then the sign of the term also changes. Positive term becomes negative and negative term becomes positive.
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