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The problem asks to convert polar equations into Cartesian coordinates. Let's address both parts:

### Part (a) \( r \cos \theta = 6 \):

In polar coordinates, we know:
\[
r \cos \theta = x
\]
Thus, the equation becomes:
\[
x = 6
\]
This is the Cartesian form of the equation, which represents a vertical line at \( x = 6 \).

### Part (b) \( r^2 - 8r \cos \theta - 4r \sin \theta + 11 = 0 \):

Using the following polar to Cartesian conversions:
\[
r^2 = x^2 + y^2, \quad r \cos \theta = x, \quad r \sin \theta = y
\]
We substitute these into the given equation:
\[
r^2 - 8r \cos \theta - 4r \sin \theta + 11 = 0
\]
becomes
\[
(x^2 + y^2) - 8x - 4y + 11 = 0
\]
This is the Cartesian form of the given equation.

Would you like further clarification or details on these steps? 

Here are 5 questions to extend this discussion:
1. How do we derive the polar to Cartesian coordinate conversion formulas?
2. Can you visualize the geometric meaning of these equations in Cartesian coordinates?
3. What is the relationship between radial distance \( r \) and \( \theta \) in polar coordinates?
4. How can we apply these conversion techniques to more complex polar equations?
5. How would the equation change if we shift or rotate the coordinate system?

**Tip:** When converting polar to Cartesian equations, remember the key conversions for \( r \) and \( \theta \): \( r^2 = x^2 + y^2 \), \( r \cos \theta = x \), and \( r \sin \theta = y \).

Ubah ke sistem koordinat kartesius, persamaan: a. r cos θ = 6, b. r^2 - 8r cos θ - 4r sin θ + 11 = 0

Learn how to convert polar equations into Cartesian coordinates using key conversion formulas. Solve equations like r cos θ = 6 and r^2 - 8r cos θ - 4r sin θ + 11 = 0 with detailed steps. Suitable for high school students in Grades 10-12.

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