Convert the following equation into an equation in terms of Cartesian coordinates.
𝑟 = 4 cos 𝜃 + 4 sin �

The given equation is in polar coordinates: 

\[
r = 4 \cos\theta + 4 \sin\theta
\]

To convert it to Cartesian coordinates, recall the relationships between polar and Cartesian coordinates:

\[
x = r \cos\theta, \quad y = r \sin\theta, \quad r^2 = x^2 + y^2
\]

### Step-by-step solution:
1. Multiply through by \(r\) (since \(r > 0\)):
   \[
   r^2 = 4r\cos\theta + 4r\sin\theta
   \]

2. Replace \(r^2\) with \(x^2 + y^2\), \(r\cos\theta\) with \(x\), and \(r\sin\theta\) with \(y\):
   \[
   x^2 + y^2 = 4x + 4y
   \]

3. Rearrange to standard Cartesian form:
   \[
   x^2 + y^2 - 4x - 4y = 0
   \]

4. Complete the square for both \(x\) and \(y\):
   - For \(x^2 - 4x\): Add and subtract \(\left(\frac{4}{2}\right)^2 = 4\)
   - For \(y^2 - 4y\): Add and subtract \(\left(\frac{4}{2}\right)^2 = 4\)

   \[
   (x^2 - 4x + 4) + (y^2 - 4y + 4) = 0 + 4 + 4
   \]

5. Simplify to:
   \[
   (x - 2)^2 + (y - 2)^2 = 8
   \]

### Final Answer:
The Cartesian equation is:

\[
(x - 2)^2 + (y - 2)^2 = 8
\]

This represents a circle with center \((2, 2)\) and radius \(\sqrt{8} = 2\sqrt{2}\).

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Would you like to see further details or have any questions? Here are 5 related questions to explore further:  
1. How can you derive the general form of a polar equation in Cartesian coordinates?  
2. What is the geometric meaning of the circle's center and radius in this case?  
3. Can you derive the equation back into polar coordinates?  
4. How does completing the square help in identifying conic sections?  
5. What are the key differences between polar and Cartesian coordinate systems?

**Tip:** Always verify your conversion by substituting sample points to ensure accuracy!

Convert the following equation into an equation in terms of Cartesian coordinates. r = 4 cos θ + 4 sin θ

Discover how to convert the polar equation r = 4 cos θ + 4 sin θ into Cartesian coordinates. This step-by-step guide includes key concepts such as polar-to-Cartesian transformation, completing the square, and identifying the resulting geometric shape—a circle with center (2, 2) and radius 2√2. Ideal for Grades 10-12 students.

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