Question

If \( A \) and \( B \) are the points of intersection of the circle \( x^2 + y^2 - 8x = 0 \) and the hyperbola \( \frac{x^2}{9} - \frac{y^2}{4} = 1 \), and a point \( P \) moves on the line \( 2x - 3y + 4 = 0 \), then the centroid of \( \triangle PAB \) lies on the line:

• A. \( 4x - 9y = 12 \)
• B. \( x + 9y = 36 \)
• C. \( 9x - 9y = 32 \)
• D. \( 6x - 9y = 20 \)

Hint:

When solving for the centroid of a triangle formed by points of intersection, use the average of the coordinates of the vertices.

Answer 1

The Correct Option is D

Step 1: Solve the System of Equations

We are given the equations of the circle and the hyperbola: \[ x^2 + y^2 - 8x = 0 \quad {(1)} \] \[ \frac{x^2}{9} - \frac{y^2}{4} = 1 \quad {(2)} \]

Completing the square in equation (1): \[ (x^2 - 8x + 16) + y^2 = 16 \quad \Rightarrow \quad (x - 4)^2 + y^2 = 16 \] This represents a circle centered at \( (4, 0) \) with radius 4.

Now multiply equation (2) by 36: \[ 4x^2 - 9y^2 = 36 \] Rearranging, \[ 13x^2 - 72x - 36 = 0 \] Solving this equation gives: \[ x = 6 \quad \text{(the valid root)} \]

Substituting this back into the circle's equation: \[ y^2 = 12 \quad \Rightarrow \quad y = \pm \sqrt{12} \] Thus, the points of intersection are: \[ A(6, \sqrt{12}) \quad \text{and} \quad B(6, -\sqrt{12}) \]

Step 2: Calculate the Centroid

The centroid of \( \triangle PAB \) is given by the average of the coordinates of \( P \), \( A \), and \( B \). Since the coordinates of \( P \) lie on the line: \[ 2x - 3y + 4 = 0 \] The centroid condition is derived as: \[ 6x - 9y = 20 \] Thus, the centroid lies on the line: \[ 6x - 9y = 20 \]