7. In the Cartesian coordinate system, if the circle with center $(r+1,0)$ and radius $r$ has a point $(a, b)$ satisfying $b^{2} \geqslant 4 a$, then the minimum value of $r$ is $\qquad$

In the Cartesian coordinate system, if the circle with center $(r+1,0)$ and radius $r$ has a point $(a, b)$ satisfying $b^{2} \geqslant 4 a$, then the minimum value of $r$ is $\qquad$
$$
\begin{array}{l}
\text { From }(a-r-1)^{2}+b^{2}=r^{2} \Rightarrow a^{2}-2(r+1) a+2 r+1+b^{2}=0 \\
\Rightarrow a^{2}-2(r+1) a+2 r+1+4 a \leqslant 0 \Rightarrow a^{2}-2(r-1) a+2 r+1 \leqslant 0,
\end{array}
$$

Notice that the above inequality has a solution, thus $\Delta=4(r-1)^{2}-4(2 r+1) \geqslant 0$ $\Rightarrow(r-1)^{2} \geqslant 2 r+1 \Rightarrow r \geqslant 4$. When the equality holds, $(a, b)=(3, \pm 2 \sqrt{3})$.