7. In the Cartesian coordinate system, if the circle with center $(r+1,0)$ and radius $r$ has a point $(a, b)$ satisfying $b^{2} \geq 4 a$, then the minimum value of $r$ is $\qquad$ .
Answer: 4.
Solution: From the condition, we have $(a-r-1)^{2}+b^{2}=r^{2}$, hence
$$
4 a \leq b^{2}=r^{2}-(a-r-1)^{2}=2 r(a-1)-(a-1)^{2}.
$$

That is, $a^{2}-2(r-1) a+2 r+1 \leq 0$.
The above quadratic inequality in $a$ has a solution, so the discriminant
$$
(2(r-1))^{2}-4(2 r+1)=4 r(r-4) \geq 0 \text{, }
$$

Solving this, we get $r \geq 4$.
Upon verification, when $r=4$, $(a, b)=(3,2 \sqrt{3})$ satisfies the condition. Therefore, the minimum value of $r$ is 4.