To solve the problem, we need to find a positive value of \( \alpha \) such that the center of the circle defined by the equation \[ x^2 + y^2 - 2\alpha x + 4y - \alpha = 0 \] is at the point \( (\alpha, -2) \) and the radius is 4 units. ### Step 1: Rewrite the Circle Equation The general form of a circle's equation is: \[ (x - h)^2 + (y - k)^2 = r^2 \] where \( (h, k) \) is the center and \( r \) is the radius. We need to rearrange the given equation into this form. Starting with: \[ x^2 + y^2 - 2\alpha x + 4y - \alpha = 0 \] we can group the \( x \) and \( y \) terms: \[ x^2 - 2\alpha x + y^2 + 4y = \alpha \] ### Step 2: Complete the Square Next, we complete the square for both \( x \) and \( y \). **For \( x \):** 1. Take the coefficient of \( x \), which is \(-2\alpha\), halve it to get \(-\alpha\), and square it to get \( \alpha^2 \). 2. Rewrite \( x^2 - 2\alpha x \) as: \[ (x - \alpha)^2 - \alpha^2 \] **For \( y \):** 1. Take the coefficient of \( y \), which is \(4\), halve it to get \(2\), and square it to get \(4\). 2. Rewrite \( y^2 + 4y \) as: \[ (y + 2)^2 - 4 \] ### Step 3: Substitute Back into the Equation Now substitute these completed squares back into the equation: \[ (x - \alpha)^2 - \alpha^2 + (y + 2)^2 - 4 = \alpha \] This simplifies to: \[ (x - \alpha)^2 + (y + 2)^2 = \alpha + \alpha^2 + 4 \] ### Step 4: Identify the Center and Radius From the rewritten equation, we can identify: - The center of the circle is \( (\alpha, -2) \). - The radius squared is \( \alpha + \alpha^2 + 4 \). ### Step 5: Set the Radius Equal to 4 Since we know the radius is 4, we set up the equation: \[ \sqrt{\alpha + \alpha^2 + 4} = 4 \] Squaring both sides gives: \[ \alpha + \alpha^2 + 4 = 16 \] ### Step 6: Solve for \( \alpha \) Rearranging this equation leads to: \[ \alpha^2 + \alpha + 4 - 16 = 0 \] which simplifies to: \[ \alpha^2 + \alpha - 12 = 0 \] ### Step 7: Factor the Quadratic Now we can factor the quadratic: \[ (\alpha - 3)(\alpha + 4) = 0 \] Setting each factor to zero gives: 1. \( \alpha - 3 = 0 \) → \( \alpha = 3 \) 2. \( \alpha + 4 = 0 \) → \( \alpha = -4 \) (not a positive value) ### Conclusion The only positive solution is \( \alpha = 3 \). ### Final Answer Thus, the correct option is **C. 3**. ### Explanation of Other Options - **A. 1**: If \( \alpha = 1 \), the radius would be \( 1 + 1^2 + 4 = 6 \), which does not equal 16. - **B. 2**: If \( \alpha = 2 \), the radius would be \( 2 + 2^2 + 4 = 10 \), which does not equal 16. - **D. 4**: If \( \alpha = 4 \), the radius would be \( 4 + 4^2 + 4 = 24 \), which does not equal 16. ### Revision Summary - The center of the circle is determined by completing the square. - The radius is derived from the rearranged circle equation. - Set the radius equal to the known value to find \( \alpha \). - The only positive solution for \( \alpha \) is 3.