To rewrite the polar equation $r = 6\cos(\theta)$ as a Cartesian equation, we use the relationships between polar and Cartesian coordinates:

1. $x = r\cos(\theta)$
2. $y = r\sin(\theta)$
3. $r^2 = x^2 + y^2$

Step-by-step Process:

1. Multiply both sides of the polar equation $r = 6\cos(\theta)$ by $r$: $r^2 = 6r\cos(\theta)$

2. Substitute $r^2 = x^2 + y^2$ and $r\cos(\theta) = x$: $x^2 + y^2 = 6x$

3. Rearrange the equation into standard form: $x^2 + y^2 - 6x = 0$

4. Complete the square for the $x$-terms: $x^2 - 6x \quad \text{can be written as} \quad (x - 3)^2 - 9.$ So, the equation becomes: $(x - 3)^2 - 9 + y^2 = 0$

5. Simplify: $(x - 3)^2 + y^2 = 9$

Final Cartesian Equation:

$(x - 3)^2 + y^2 = 9$

This represents a circle with center at $(3, 0)$ and radius $3$.

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Do you want a detailed explanation of any step?
Here are related questions:

1. How do you complete the square in a quadratic equation?
2. What is the geometric interpretation of the equation $(x - 3)^2 + y^2 = 9$?
3. How do the polar-to-Cartesian transformations work in general?
4. How do you find the radius and center of a circle from its equation?
5. Can this equation represent anything other than a circle in certain conditions?

Tip: Remember that completing the square is a key skill for converting quadratic equations into standard forms!