Key Concepts and Formulas

• Centroid of a Triangle: The centroid of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ has coordinates $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$.
• Equation of a Circle: The equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$. The general form is $x^2 + y^2 + 2gx + 2fy + c = 0$.
• Equation of a Hyperbola: The standard equation of a hyperbola centered at the origin is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Step-by-Step Solution

Step 1: Find the intersection points A and B of the circle and hyperbola.

We need to find the coordinates of the intersection points of the circle $x^2 + y^2 - 8x = 0$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$.

1. Rewrite the equations:

   • Circle: $x^2 + y^2 - 8x = 0 \quad \ldots (1)$
   • Hyperbola: $\frac{x^2}{9} - \frac{y^2}{4} = 1 \Rightarrow 4x^2 - 9y^2 = 36 \quad \ldots (2)$

2. Solve simultaneously: From equation (1), $y^2 = 8x - x^2$. Substitute this into equation (2): $4x^2 - 9(8x - x^2) = 36$ $4x^2 - 72x + 9x^2 = 36$ $13x^2 - 72x - 36 = 0 \quad \ldots (3)$

3. Solve the quadratic equation for x: Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{72 \pm \sqrt{(-72)^2 - 4(13)(-36)}}{2(13)}$ $x = \frac{72 \pm \sqrt{5184 + 1872}}{26}$ $x = \frac{72 \pm \sqrt{7056}}{26}$ $x = \frac{72 \pm 84}{26}$ The two possible x-coordinates are:

   • $x_1 = \frac{72 + 84}{26} = \frac{156}{26} = 6$
   • $x_2 = \frac{72 - 84}{26} = \frac{-12}{26} = -\frac{6}{13}$

4. Check for valid x-coordinates: For the circle, $y^2 = 8x - x^2 \ge 0$, which implies $x(8-x) \ge 0$, so $0 \le x \le 8$. For the hyperbola, $y^2 = \frac{4x^2 - 36}{9} \ge 0$, which implies $x^2 \ge 9$, so $x \le -3$ or $x \ge 3$. Combining these, we need $3 \le x \le 8$. $x_1 = 6$ is a valid x-coordinate since $3 \le 6 \le 8$. $x_2 = -\frac{6}{13}$ is not a valid x-coordinate.

5. Find the corresponding y-coordinates: Substitute $x = 6$ into the circle equation (1): $6^2 + y^2 - 8(6) = 0$ $36 + y^2 - 48 = 0$ $y^2 = 12$ $y = \pm \sqrt{12} = \pm 2\sqrt{3}$

   Thus, the points of intersection are $A = (6, 2\sqrt{3})$ and $B = (6, -2\sqrt{3})$.

Step 2: Calculate the sum of the x and y coordinates of A and B.

This step simplifies the centroid calculation.

• $x_A + x_B = 6 + 6 = 12$
• $y_A + y_B = 2\sqrt{3} - 2\sqrt{3} = 0$

Step 3: Define the coordinates of the moving point P.

Let the coordinates of point P be $(x_P, y_P)$. Since P moves on the line $2x - 3y + 4 = 0$, we have $2x_P - 3y_P + 4 = 0$.

Step 4: Express the coordinates of the centroid G in terms of the coordinates of A, B, and P.

Let $G(h, k)$ be the centroid of $\Delta PAB$. Using the centroid formula: $h = \frac{x_A + x_B + x_P}{3} = \frac{12 + x_P}{3} \quad \ldots (4)$ $k = \frac{y_A + y_B + y_P}{3} = \frac{0 + y_P}{3} = \frac{y_P}{3} \quad \ldots (5)$

Step 5: Eliminate $x_P$ and $y_P$ to find the locus of the centroid.

We need to find a relationship between $h$ and $k$ that is independent of $x_P$ and $y_P$. From equations (4) and (5), we can express $x_P$ and $y_P$ in terms of $h$ and $k$:

• From (4): $3h = 12 + x_P \implies x_P = 3h - 12$
• From (5): $3k = y_P \implies y_P = 3k$

Substitute these expressions for $x_P$ and $y_P$ into the equation of the line on which P moves ($2x_P - 3y_P + 4 = 0$): $2(3h - 12) - 3(3k) + 4 = 0$ $6h - 24 - 9k + 4 = 0$ $6h - 9k - 20 = 0$ $6h - 9k = 20$

Step 6: Write the equation of the locus of the centroid.

Replacing $(h, k)$ with $(x, y)$ to represent the general coordinates of the centroid, the locus of the centroid of $\Delta PAB$ is: $6x - 9y = 20$

Step 7: Compare with the given options.

The equation of the locus of the centroid is $6x - 9y = 20$. Comparing with the given options, we see that this matches option (D).

Common Mistakes & Tips

• Domain Check: Always check for the valid ranges of $x$ and $y$ when dealing with intersection points of curves. Real intersection points must satisfy the conditions for both curves.
• Centroid Locus Property: If two vertices are fixed and the third moves on a line, the centroid's locus is a line parallel to the original line.
• Algebraic Manipulation: Be meticulous with algebraic manipulations and substitutions to avoid errors.

Summary

We found the intersection points A and B of the circle and hyperbola. Then, using the centroid formula and the equation of the line on which P moves, we expressed the coordinates of the centroid in terms of P's coordinates. Finally, we eliminated the coordinates of P to find the locus of the centroid, which is the line $6x - 9y = 20$.

The final answer is \boxed{6x - 9y = 20}, which corresponds to option (D).