, 2024

Solution:

Invert about the unit circle centered at the origin. $\omega$ turns into a horizontal line, and the hyperbola turns into the following:
$$\begin{aligned} \frac{x^{2}}{\left(x^{2}+y^{2}\right)^{2}}-\frac{20 y^{2}}{\left(x^{2}+y^{2}\right)^{2}}=24 & \Longrightarrow x^{2}-20 y^{2}=24\left(x^{2}+y^{2}\right)^{2} . \\ & \Longrightarrow 24 x^{4}+\left(48 y^{2}-1\right) x^{2}+24 y^{4}+20 y^{2}=0 \\ & \Longrightarrow\left(48 y^{2}-1\right)^{2} \geq 4(24)\left(24 y^{4}+20 y^{2}\right) \\ & \Longrightarrow 1-96 y^{2} \geq 1920 y^{2} \\ & \Longrightarrow y \leq \sqrt{1 / 2016} . \end{aligned}$$
This means that the horizontal line in question is $y=\sqrt{1 / 2016}$. This means that the diameter of the circle is the reciprocal of the distance between the point and line, which is $\sqrt{2016}$, so the radius is $\sqrt{504}$, and the answer is $504 \pi$.

Solution:
Let $a$ be the $y$-coordinate of both tangency points to the hyperbola. Then, the equation of the circle must be in the form
$$x^{2}-20 y^{2}+c(y-a)^{2}=24 .$$
Comparing the $y^{2}$-coefficient, we see that $c=21$. Moreover, we need it to pass through $(0,0)$, so $21 a^{2}=24$. Thus, the equation of the circle is
$$x^{2}+y^{2}-42 a y+21 a^{2}=24 \Longrightarrow x^{2}+(y-21 a)^{2}=(21 a)^{2},$$
so the radius is $21 a$, and the area is $\left(441 a^{2}\right) \pi=504 \pi$.