JEE Main 2025 (Online) 28th January Evening ShiftQuestion 1
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 :
Hyperbola
Solution
Subtracting, $x^2+y^2=8x$ and $4x^2-9y^2=36$ give $13x^2-72x-36=0$, so $x=6$ and $A,B=(6,\pm2\sqrt3)$. Centroid $G=\left(\frac{h+12}3,\frac k3\right)$; substituting $h=3X-12$, $k=3Y$ in $2h-3k+4=0$ gives $6X-9Y-20=0$.