Question

Consider the point \( P(\alpha, \beta) \) on the line \( 2x + y = 1 \). If the points \( P \) and \( (3,2) \) are conjugate points with respect to the circle \( x^2 + y^2 = 4 \), then find \( \alpha + \beta \):

• A. \( 3 \)
• B. \( -1 \)
• C. \( -5 \)
• D. \( 7 \)

Hint:

Use the conjugate points property to form an equation and solve along with the given line equation.

Answer 1

The Correct Option is A

Step 1: Conjugate points condition Two points \( (x_1, y_1) \) and \( (x_2, y_2) \) are conjugates w.r.t. a circle \( x^2 + y^2 = r^2 \) if: \[ x_1 x_2 + y_1 y_2 = r^2. \] For the given circle \( x^2 + y^2 = 4 \), the equation becomes: \[ 3\alpha + 2\beta = 4. \] Step 2: Find \( \alpha \) and \( \beta \) Since \( P(\alpha, \beta) \) lies on the line \( 2x + y = 1 \), we substitute: \[ 2\alpha + \beta = 1. \] Step 3: Solve for \( \alpha + \beta \) Solving the system: \[ 3\alpha + 2\beta = 4, \quad 2\alpha + \beta = 1. \] Multiplying the second equation by 2: \[ 4\alpha + 2\beta = 2. \] Subtracting, \[ (3\alpha + 2\beta)
- (4\alpha + 2\beta) = 4
- 2. \] \[
-\alpha = 2 \Rightarrow \alpha =
-2. \] Substituting in \( 2\alpha + \beta = 1 \): \[ 2(
-2) + \beta = 1 \Rightarrow
-4 + \beta = 1 \Rightarrow \beta = 5. \] \[ \alpha + \beta =
-2 + 5 = 3. \]