If $A$ and $B$ are the points of intersection of the circle $x^2 + y^2 - 8x = 0$ and the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{4} = 1$ and a point $P$ moves on the line $2x - 3y + 4 = 0$, then the centroid of $\triangle PAB$ lies on the line
Step-by-Step Solution
Key Concept: Find the fixed points $A$ and $B$ by solving the circle and hyperbola simultaneously; express the centroid $G=(\tfrac{x_A+x_B+h}{3}, \tfrac{y_A+y_B+k}{3})$ where $(h,k)$ lies on the given line; eliminate $h$ and $k$ to get the locus of $G$.
From the circle: $y^2=8x-x^2$. Substituting into the hyperbola: $\tfrac{x^2}{9}-\tfrac{8x-x^2}{4}=1 \Rightarrow 13x^2-72x-36=0 \Rightarrow x=6$ or $x=-\tfrac{13}{6}$ (rejected as it lies outside the circle). $y^2=8(6)-36=12$, so $A=(6,2\sqrt{3})$, $B=(6,-2\sqrt{3})$. $P=(h,\tfrac{2h+4}{3})$ on $2x-3y+4=0$. Centroid $G=\left(\tfrac{12+h}{3},\tfrac{2h+4}{9}\right)$. Let $G=(X,Y)$: $h=3X-12$, $Y=\tfrac{6X-20}{9}$, i.e., $9Y=6X-20$, or $6X-9Y=20$.
Correct Answer: 3