Question

The circle $x^2+y^2-8x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at point $A$ and $B$. The equation of circle with $AB$ as diameter is:

• A. $x^2+y^2-12x+24=0$
• B. $x^2+y^2+12x+24=0$
• C. $x^2+y^2+29x-12=0$
• D. $x^2+y^2-24x-12=0$

Answer 1

The correct option is A $x^2+y^2-12x+24=0$

A point on hyperbola is $(3\sec \theta ,2\tan \theta )$
This point lies on the circle, so
$9{\sec }^2\theta +4{\tan }^2\theta -24\sec \theta =0\Rightarrow 13{\sec }^2\theta -24\sec \theta -4=0\Rightarrow (13\sec \theta +2)(\sec \theta -2)=0\Rightarrow \sec \theta =2,-\frac{2}{13}\therefore \sec \theta =2\Rightarrow \tan \theta =\pm \sqrt{3}$

So, the point of intersection are
$A(6,2\sqrt{3})\text{and}B(6,-2\sqrt{3})$
$\therefore$ The circle equation with $AB$ as diameter is
$(x-6{)}^2+y^2=(2\sqrt{3}{)}^2\Rightarrow x^2+y^2-12x+24=0$