HMMT-Nov 2023 guts, Problem 23

The points $A=\left(4, \frac{1}{4}\right)$ and $B=\left(-5,-\frac{1}{5}\right)$ lie on the hyperbola $x y=1$. The circle with diameter $A B$ intersects this hyperbola again at points $X$ and $Y$. Compute $X Y$.

Solution: \section*{Solution:} Let $A=(a, 1 / a), B=(b, 1 / b)$, and $X=(x, 1 / x)$. Since $X$ lies on the circle with diameter $\overline{A B}$, we have $\angle A X B=90^{\circ}$. Thus, $\overline{A X}$ and $\overline{B X}$ are perpendicular, and so the product of their slopes must be -1 . We deduce: \[ \frac{a-x}{\frac{1}{a}-\frac{1}{x}} \frac{b-x}{\frac{1}{b}-\frac{1}{x}}=-1 \Longrightarrow(a x)(b x)=-1 \] so $x= \pm \sqrt{-a b}$. Plugging in $a=4$ and $b=-5$ gives $X=(\sqrt{20}, 1 / \sqrt{20})$ and $Y=(-\sqrt{20},-1 / \sqrt{20})$, giving the answer. $\fbox{\sqrt{\frac{401}{5}}}$.

Answer: \sqrt{\frac{401}{5}}