Question

The center and radius of the circle \(r^2 = 4r \cos \theta\) are:

• A. Center = (4,0), Radius = 4
• B. Center = (0,2), Radius = 2
• C. Center = (0,2), Radius = 4
• D. Center = (2,0), Radius = 2

Hint:

Exam Tip:
When converting polar to Cartesian:

• Use \(x = r \cos \theta\), \(y = r \sin \theta\).
• Use \(r^2 = x^2 + y^2\).
• Multiply by \(r\) if necessary to eliminate \(r\) in denominators.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question involves converting a polar equation to Cartesian coordinates to identify the center and radius of the circle.

Step 2: Key Formula or Approach:
The polar coordinates are \(x = r \cos \theta\), \(y = r \sin \theta\), and \(r^2 = x^2 + y^2\).
We need to convert \(r^2 = 4r \cos \theta\) into Cartesian form.

Step 3: Detailed Explanation:
Given \(r^2 = 4r \cos \theta\).
Divide both sides by \(r\) (assuming \(r \neq 0\)): \[ r = 4 \cos \theta \] Multiply both sides by \(r\): \[ r^2 = 4r \cos \theta \] Substitute \(r^2 = x^2 + y^2\) and \(r \cos \theta = x\): \[ x^2 + y^2 = 4x \] Complete the square for \(x\): \[ x^2 - 4x + y^2 = 0 \quad \Rightarrow \quad (x^2 - 4x + 4) + y^2 = 4 \] \[ (x - 2)^2 + y^2 = 4 \] This is the equation of a circle with center \((2, 0)\) and radius \(2\) (since \(r^2 = 4\), so \(r = 2\)).

Step 4: Final Answer:
Center = (2, 0), Radius = 2. Therefore, option (D) is correct.