Question:

<p>Given equation of circle<br /> <span class="math-tex">$x^{2}+y^{2}-8 x=0$</span>, and hyperbola<br /> <span class="math-tex">$\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$</span>&nbsp;...(i)<br /> <span class="math-tex">$4 x^{2}-9 y^{2}=36$</span>&nbsp;...(ii)<br /> Solving equation (i) and (ii)<br /> <span class="math-tex">$\Rightarrow 4 x^{2}-9\left(8 x-x^{2}\right)=36$</span><br /> <span class="math-tex">$\Rightarrow 13 x^{2}-72 x-36=0$</span><br /> <span class="math-tex">$\Rightarrow(13 x+6)(x-6)=0$</span><br /> <span class="math-tex">$\therefore x=\frac{-6}{13}, 6$</span><br /> <span class="math-tex">$x=\frac{-6}{13}$</span> rejected <span class="math-tex">$y \rightarrow$</span> Imaginary<br /> When,<br /> <span class="math-tex">$x=6, \frac{36}{9}-\frac{y^{2}}{4}=$</span><span class="math-tex">$1 \Rightarrow \frac{y^{2}}{4}=\frac{36-9}{9}$</span><br /> <span class="math-tex">$\Rightarrow y^{2}=12 \Rightarrow y= \pm \sqrt{12}$</span><br /> <span class="math-tex">$A(6, \sqrt{12}), B(6,-\sqrt{12})$</span><br /> <span class="math-tex">${p}\left(\alpha, \frac{2 \alpha+4}{3}\right)$</span> centroid <span class="math-tex">$(h, k)$</span><br /> <span class="math-tex">$P$</span> lies on <span class="math-tex">$2 x-3 y+4=0$</span><br /> <span class="math-tex">$\therefore P$</span> lies on the line <span class="math-tex">$2 x-3 y+4=0$</span><br /> Let centroid ( <span class="math-tex">$h, k$</span> )<br /> <span class="math-tex">$\therefore(h, k)=\left(\frac{6+6+\alpha}{3}, \frac{\sqrt{12}-\sqrt{12}+\frac{2 \alpha+4}{3}}{3}\right)$</span><br /> <span class="math-tex">$\therefore h=\frac{12+\alpha}{3} \Rightarrow \alpha=3 h-12$</span><br /> and <span class="math-tex">$k=\frac{\frac{2 \alpha+4}{3}}{3} \Rightarrow 2 \alpha+4=9 k$</span><br /> <span class="math-tex">$\Rightarrow \alpha=\frac{9 k-4}{2}$</span><br /> <span class="math-tex">$6 h-2 y=9 k-4$</span><br /> Hence, <span class="math-tex">$6 x-9 y=20$</span></p>