Question

Let $x^2+y^2=4r^2$ and $xy=1$ intersects at $A$ and $B$ in first quadrant. If $AB=\sqrt{14}$ units, then the value of $\left|2r\right|$ is

Answer 1

Answer: 3

Let $A=t,\frac{1}{t}\&B=\frac{1}{t},t$
$\because$ Both curves are symmetric about line $y=x$
then, $A{B}^2=t-{\frac{1}{t}}^2+\frac{1}{t}-t^2=14$
$\Rightarrow 2t^2+\frac{1}{t^2}=14+4$
$\Rightarrow t^2+\frac{1}{t^2}=9$
Now, $O{A}^2=t^2+\frac{1}{t^2}=9\Rightarrow 2r^2=9$
$\Rightarrow 2r=3$