Mathematics - Circle Question with Solution | TestHub

The equation of the hyperbola is $\frac{x^2}{9}-\frac{y^2}{4}=1$ and that of circle is $$x^2+y^2-8x=0$$ For their points of intersection $$\frac{x^2}{9}+\frac{x^2-8x}{4}=1$$ $$\begin{array}{ll}\Rightarrow & 4x^2+9x^2-72x=36\\ \Rightarrow & 13x^2-72x-36=0\end{array}$$ $$\begin{array}{rl}& \Rightarrow 13x^2-78x+6x-36=0\\ & \Rightarrow 13x(x-6)+6(x-6)=0\\ & \Rightarrow x=6,x=-\frac{13}{6}\\ & x=-\frac{13}{6}\text{ not acceptable }\\ & \end{array}$$ Now, for $x=6,y=\pm 2\sqrt{3}$ Required equation is $$\begin{array}{rl}& (x-6{)}^2+(y+2\sqrt{3})(y-2\sqrt{3})=0\\ & \Rightarrow x^2-12x+y^2+24=0\\ & \Rightarrow x^2+y^2-12x+24=0\end{array}$$