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 $B$.

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

Given: $x^2+y^2-8x=0.....\left(1\right)$ $⟹y^2=8x-x^2$

$\frac{x^2}{9}-\frac{y^2}{4}\Rightarrow 1$

$\Rightarrow \frac{x^2}{9}-\frac{(8x-x^2)}{4}=1$

$\Rightarrow 4x^2-9(8x-x^2)=36$

$\Rightarrow 4x^2-72x+9x^2-36=0$

$\Rightarrow 13x^2-72x-36=0$

$\Rightarrow x=\frac{72\pm \sqrt{(72{)}^2-4\times 13\times 36}}{26}$

$\Rightarrow x=6$

Then, putting $x=6$ in equation (1), we get

$\Rightarrow y=\pm \sqrt{12}=\pm \sqrt{4\times 3}=\pm 2\sqrt{3}$

So, radius $=2\sqrt{3}$

As $AB$ is a diameter center will be the midpoint of point $A=(6,2\sqrt{3})$ and $B=(6,-2\sqrt{3})$

By mid point formula,

Center $=(\frac{6+6}{2},\frac{2\sqrt{3}-2\sqrt{3}}{2})$

Centre $=(6,0)$

$\Rightarrow (x-6{)}^2+y^2=12$

$\Rightarrow x^2+36-12x+y^2=12$

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