10. (This question is worth 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 $\Gamma$. Find the radius of circle $\Omega$.
Solution: It is easy to know that the coordinates of the focus $F$ of $\Gamma$ are $(1,0)$. Let the radius of circle $\Omega$ be $r(r>0)$. By symmetry, without loss of generality, assume that $\Omega$ is tangent to the $x$-axis at $F$ above the $x$-axis, so the equation of $\Omega$ is
$$
(x-1)^{2}+(y-r)^{2}=r^{2} \text {. }
$$

Substituting $x=\frac{y^{2}}{4}$ into (1) and simplifying, we get $\left(\frac{y^{2}}{4}-1\right)^{2}+y^{2}-2 r y=0$. Clearly $y>0$, hence
$$
r=\frac{1}{2 y}\left(\left(\frac{y^{2}}{4}-1\right)^{2}+y^{2}\right)=\frac{\left(y^{2}+4\right)^{2}}{32 y} \text {. }
$$
.................. 5 points
According to the condition, (2) has exactly one positive solution $y$, which corresponds to the unique common point of $\Omega$ and $\Gamma$.
Consider the minimum value of $f(y)=\frac{\left(y^{2}+4\right)^{2}}{32 y}(y>0)$.
By the AM-GM inequality, we know $y^{2}+4=y^{2}+\frac{4}{3}+\frac{4}{3}+\frac{4}{3} \geq 4 \sqrt[4]{y^{2} \cdot\left(\frac{4}{3}\right)^{3}}$, thus
$$
f(y) \geq \frac{1}{32 y} \cdot 16 \sqrt{y^{2} \cdot\left(\frac{4}{3}\right)^{3}}=\frac{4 \sqrt{3}}{9} .
$$

When and only when $y^{2}=\frac{4}{3}$, i.e., $y=\frac{2 \sqrt{3}}{3}$, $f(y)$ reaches its minimum value $\frac{4 \sqrt{3}}{9} . \cdots \cdots \cdots \cdots \cdots \cdots \cdot 15$ points
From (2) having a solution, we know $r \geq \frac{4 \sqrt{3}}{9}$. If $r>\frac{4 \sqrt{3}}{9}$, since $f(y)$ varies continuously with $y$, and $f(y)$ can be arbitrarily large as $y \rightarrow 0^{+}$ and $y \rightarrow+\infty$, (2) has solutions in $\left(0, \frac{2 \sqrt{3}}{3}\right)$ and $\left(\frac{2 \sqrt{3}}{3},+\infty\right)$, which contradicts the uniqueness of the solution.
In summary, only $r=\frac{4 \sqrt{3}}{9}$ satisfies the condition (at this time, $\left(\frac{1}{3}, \frac{2 \sqrt{3}}{3}\right)$ is the unique common point of $\Omega$ and $\Gamma$).
20 points