10. 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 $\Omega$.
Assume that circle $\Omega$ is above the $x$-axis, then $\Omega: (x-1)^{2}+(y-r)^{2}=r^{2}$, i.e., $x^{2}-2 x+1+y^{2}-2 r y=0$. Substituting $y^{2}=4 x$ yields $\frac{y^{4}}{16}+\frac{y^{2}}{2}+1-2 r y=0 \Rightarrow r=\frac{y^{4}+8 y^{2}+16}{32 y}=\frac{\left(y^{2}+4\right)^{2}}{32 y}(y>0)$.
According to the problem, $r=\frac{\left(y^{2}+4\right)^{2}}{32 y}(y>0)$ has a unique solution for $y$. By the AM-GM inequality, we get $y^{2}+4=y^{2}+\frac{4}{3}+\frac{4}{3}+\frac{4}{3} \geqslant 4 \sqrt[4]{\left(\frac{4}{3}\right)^{3} y^{2}} \Rightarrow\left(y^{2}+4\right)^{2} \geqslant 16 \sqrt{\left(\frac{4}{3}\right)^{3} y^{2}}$ $=\frac{16 \times 8 \sqrt{3}}{9} y \Rightarrow r \geqslant \frac{4 \sqrt{3}}{9}$, equality holds when $y^{2}=\frac{4}{3} \Rightarrow y=\frac{2 \sqrt{3}}{3}$. Noting that the function $r=\frac{\left(y^{2}+4\right)^{2}}{32 y}(y>0)$ approaches $+\infty$ as $y \rightarrow 0$ and $y \rightarrow+\infty$, thus if $r>\frac{4 \sqrt{3}}{9}$, there must be one solution each in $\left(0, \frac{2 \sqrt{3}}{3}\right)$ and $\left(\frac{2 \sqrt{3}}{3},+\infty\right)$, which is a contradiction. In summary, the radius of the circle $\Omega$ is $r=\frac{4 \sqrt{3}}{9}$.