Question

How do you convert \[r = 12\cos [\theta ] \] to rectangular form?

Answer 1

Hint: According to the question, first we will multiply the equation with ‘r’. Then we will substitute the values. Then we will try to solve ‘h’. For that we will try to make such an equation that gets the formula of \[{(a - b)^2} = {a^2} + {b^2} - 2ab\] . After getting the value of ‘h’ by solving this, we will square the terms and get the correct equation.
Formula used: \[{(a - b)^2} = {a^2} + {b^2} - 2ab\] .

Complete step-by-step answer:
The given equation is:
 \[r = 12\cos [\theta ] \]
First, we will start by multiplying ‘r’ on both the sides of the equation, and we get:
 \[r \times r = 12\cos (\theta ) \times r\]
When we solve this, we get:
 \[ \Rightarrow {r^2} = 12r\cos (\theta )\]
Here, we will try to substitute \[{x^2} + {y^2}\] for \[{r^2}\] and \[x\] for \[r\cos (\theta )\] , and we get:
 \[{x^2} + {y^2} = 12x\]
Now, we will subtract -12x form both the sides of the equation, and we get:
 \[{x^2} + {y^2} - 12x = 12x - 12x\]
After solving it, we get:
 \[ \Rightarrow {x^2} + {y^2} - 12x = 0\]
Now, we will try to add \[{h^2}\] on both the sides of the equation, and we get:
 \[{x^2} + {y^2} - 12x + {h^2} = {h^2}\]
This equation can also be written as:
 \[ \Rightarrow {x^2} + {h^2} - 12x + {y^2} = {h^2}\]
We know the formula:
 \[{(a - b)^2} = {a^2} + {b^2} - 2ab\]
We will apply this formula in the equation, and we get:
 \[{x^2} + {h^2} - 12x\]
From this, we get to know that:
 \[ - 2hx = - 12x\]
 \[ \Rightarrow h = 6\]
Now, we will try to insert the square terms with the value \[h = 6\] on the left side. On the right side, we will substitute the number 6 for the variable ‘h’. This will give us:
 \[{(x - 6)^2} + {y^2} = {6^2}\]
Now, at last, we will try to write the terms which are in ‘y’, in a standard form, and we get:
 \[{(x - 6)^2} + {(y - 0)^2} = {6^2}\]
Therefore, this is the final result.

Note: According to this question, the equation here is becoming a circle which is having a radius of 6. This is also centered at (6, 0). This question comes under the portion of converting the systems in the Polar system in the trigonometry section.