JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Circle

To find the intersection points A and B, solve the system: $x^2+y^2=8x$ and $\frac{x^2}{9}-\frac{y^2}{4}=1$. From the circle equation, $y^2=8x-x^2$. Substituting into the hyperbola equation and simplifying yields the x-coordinates of A and B.

The equation of the circle with AB as diameter can be found using the radical axis approach. The equation is of the form $(x^2+y^2-8x)+\lambda (\frac{x^2}{9}-\frac{y^2}{4}-1)=0$ for some $\lambda$. The circle passes through A and B (both satisfy the original equations), and we need the coefficient of $x^2$ and $y^2$ to be equal.

Setting coefficients equal: $1+\frac{\lambda }{9}=1-\frac{\lambda }{4}$ gives $\lambda =-\frac{36}{13}$. Substituting and simplifying yields $x^2+y^2-12x+24=0$.