Question

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

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

Answer 1

The correct option is A

$x^2+y^2-12x+24=0$

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 , $\frac{x^2}{9}+\frac{x^2-8x}{4}=1$
$\Rightarrow 4x^2+9x^2-72x=36$
$\Rightarrow 13x^2-72x-36=0$
$\Rightarrow 13x^2-78x+6x-36=0$
$\Rightarrow 13x(x-6)+6(x-6)=0$
$\Rightarrow x=6,x=-\frac{13}{6}$
$x=-\frac{13}{6}$ not acceptable.
Now, for $x=6,y=\pm 2\sqrt{3}$
Required equation is, $(x-6{)}^2+(y+2\sqrt{3}\left)\right(y-2\sqrt{3})=0$
$\Rightarrow x^2-12x+y^2+24=0$
$\Rightarrow x^2+y^2-12x+24=0$