Rewrite the polar equation r=6cos(θ) as a Cartesian **equation**.

  


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 \( 3 \).

---

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!

Rewrite the polar equation r=6cos(θ) as a Cartesian equation.

Explore how to rewrite the polar equation r=6cos(θ) as a Cartesian equation. This step-by-step guide demonstrates the conversion process, using key formulas like x = r*cos(θ), r^2 = x^2 + y^2, and completing the square. Understand how this results in a circle equation with a center at (3, 0) and radius 3. Ideal for Grades 9-12 students studying coordinate geometry.

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