Answer:
$\frac{1000}{9} \pi$
Solution: Let $P$ satisfy $O P=x$. (For now, we focus on $f(P)$ and ignore the $A$ and $B$ from the problem statement.) The key idea is that if we invert at some point along $O P$ such that the images of $\Gamma$ and $\Omega$ are concentric, then $\omega_{i}$ still exist. Suppose that this inversion fixes $\Gamma$ and takes $\Omega$ to $\Omega^{\prime}$ of radius $r$ (and $X$ to $X^{\prime}$ in general). If the inversion is centered at a point $Q$ along ray $O P$ such that $O Q=d$, then the radius of inversion is $\sqrt{d^{2}-1}$. Let the diameter of $\Omega$ meet $O Q$ at $A$ and $B$ with $A$ closer to $Q$ than $B$. Then, $\left(A B ; P P_{\infty}\right)=-1$ inverts to $\left(A^{\prime} B^{\prime} ; P^{\prime} Q\right)=-1$, where $P_{\infty}$ is the point at infinity along line $O P$, so $P^{\prime}$ is the inverse of $Q$ in $\Omega^{\prime}$. We can compute $O P^{\prime}=\frac{r^{2}}{d}$ so $P^{\prime} Q=d-\frac{r^{2}}{d}$ and $P Q=\frac{d^{2}-1}{d-\frac{r^{2}}{d}}$. Thus, we get the equation $\frac{d^{2}-1}{d-\frac{r^{2}}{d}}+x=d$, which rearranges to $\frac{1-r^{2}}{d^{2}-r^{2}} d=x$, or $d^{2}-x^{-1}\left(1-r^{2}\right) d-r^{2}=0$. Now, we note that the radius of $\Omega$ is
$$
\frac{1}{2} A B=\frac{1}{2}\left(\frac{d^{2}-1}{d-r}-\frac{d^{2}-1}{d+r}\right)=\frac{r\left(d^{2}-1\right)}{d^{2}-r^{2}}=r\left(1+\frac{r^{2}-1}{d^{2}-r^{2}}\right)=r\left(1-\frac{x}{d}\right) .
$$

The quadratic formula gives us that $d=\frac{\left(1-r^{2}\right) \pm \sqrt{r^{4}-\left(2-4 x^{2}\right) r^{2}+1}}{2 x}$, so $\frac{x}{d}=-\frac{1-r^{2} \pm \sqrt{r^{4}-\left(2-4 x^{2}\right) r^{2}+1}}{2 r^{2}}$, which means that the radius of $\Omega$ is
$$
\frac{r^{2}+1 \pm \sqrt{r^{4}-\left(2-4 x^{2}\right) r^{2}+1}}{2 r}=\frac{r+\frac{1}{r} \pm \sqrt{r^{2}+\frac{1}{r^{2}}-2+4 x^{2}}}{2} .
$$

Note that if $r$ gives a valid chain of 2020 circles, so will $\frac{1}{r}$ by homothety/inversion. Thus, we can think of each pair of $r, \frac{1}{r}$ as giving rise to two possible values of the radius of $\Omega$, which are $\frac{r+\frac{1}{r} \pm \sqrt{r^{2}+\frac{1}{r^{2}}-1}}{2}$. This means that the pairs have the same sum of radii as the circles centered at $O$, and the product of the radii is $1-x^{2}$. (A simpler way to see this is to note that inversion at $P$ with radius $\sqrt{1-x^{2}}$ swaps the two circles.) From this, it follows that the difference between the sum of the areas for each pair is $2 \pi\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5}{18} \pi$. There are $\frac{\varphi(2020)}{2}=400$ such pairs, which can be explicitly computed as
10. Let $\Gamma$ be a circle of radius 1 centered at $O$. A circle $\Omega$ is said to be friendly if there exist distinct circles $\omega_{1}, \omega_{2}, \ldots, \omega_{2020}$, such that for all $1 \leq i \leq 2020, \omega_{i}$ is tangent to $\Gamma, \Omega$, and $\omega_{i+1}$. (Here, $\omega_{2021}=\omega_{1}$.) For each point $P$ in the plane, let $f(P)$ denote the sum of the areas of all friendly circles centered at $P$. If $A$ and $B$ are points such that $O A=\frac{1}{2}$ and $O B=\frac{1}{3}$, determine $f(A)-f(B)$.