10. (20 points) In the Cartesian coordinate system $x O y$, circle $\Omega$ intersects the parabola $\Gamma: y^{2}=4 x$ at exactly one point, and circle $\Omega$ is tangent to the $x$-axis at the focus $F$ of the parabola $\Gamma$. Find the radius of circle $\Omega$.


10. It is known that the focus of the parabola $\Gamma$ is $F(1,0)$.

Let the radius of the circle $\Omega$ be $r(r>0)$.
By symmetry, we assume that the circle $\Omega$ is tangent to the $x$-axis at point $F$ above the $x$-axis. Then the equation of the circle $\Omega$ is
$$
(x-1)^{2}+(y-r)^{2}=r^{2} \text {. }
$$

Substituting $x=\frac{y^{2}}{4}$ into equation (1) and simplifying, we get
$$
\left(\frac{y^{2}}{4}-1\right)^{2}+y^{2}-2 r y=0 \text {. }
$$

Clearly, $y>0$.
Thus, $r=\frac{1}{2 y}\left(\left(\frac{y^{2}}{4}-1\right)^{2}+y^{2}\right)=\frac{\left(y^{2}+4\right)^{2}}{32 y}$.
According to the given conditions, we know that equation (2) has exactly one positive solution $y$, which corresponds to the unique common point of the circle $\Omega$ and the parabola $\Gamma$.
Consider the minimum value of $f(y)=\frac{\left(y^{2}+4\right)^{2}}{32 y}(y>0)$.
By the AM-GM inequality, we have
$y^{2}+4=y^{2}+\frac{4}{3}+\frac{4}{3}+\frac{4}{3} \geqslant 4 \sqrt[4]{y^{2}\left(\frac{4}{3}\right)^{3}}$.
Thus, $f(y) \geqslant \frac{1}{32 y} \cdot 16 \sqrt{y^{2}\left(\frac{4}{3}\right)^{3}}=\frac{4 \sqrt{3}}{9}$,

with equality if and only if $y^{2}=\frac{4}{3}$, i.e., $y=\frac{2 \sqrt{3}}{3}$, at which $f(y)$ attains its minimum value $\frac{4 \sqrt{3}}{9}$.

From the fact that equation (2) has a solution, we know that $r \geqslant \frac{4 \sqrt{3}}{9}$.
Furthermore, if $r>\frac{4 \sqrt{3}}{9}$, noting that $f(y)$ varies continuously with $y$, and that $f(y)$ can be arbitrarily large as $y \rightarrow 0^{+}$ and $y \rightarrow+\infty$, it follows that equation (2) has solutions in both $\left(0, \frac{2 \sqrt{3}}{3}\right)$ and $\left(\frac{2 \sqrt{3}}{3},+\infty\right)$, contradicting the uniqueness of the solution.

In summary, only $r=\frac{4 \sqrt{3}}{9}$ satisfies the conditions, and at this time, the unique common point of the circle $\Omega$ and the parabola $\Gamma$ is $\left(\frac{1}{3}, \frac{2 \sqrt{3}}{3}\right)$.