Example 7 The parabola $C_{1}: y=x^{2}+2 a x+b$ intersects the $x$-axis at $A, B$. If the vertex of $C_{1}$ is always inside the circle $C_{2}$ with segment $A B$ as its diameter. Try to find the relationship that $a, b$ should satisfy.
Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$. From $C_{1}$, setting $y=0$ gives $x^{2}+2 a x+b=0$. Under the condition $\Delta>0$ i.e., $a^{2}>b$, we have $x_{1}+x_{2}=-2 a, x_{1} \cdot x_{2}=b$.

The equation of circle $C_{2}$ can be set as $\left(x-x_{1}\right)\left(x-x_{2}\right)+(y-0)(y-0)=0$, simplifying and rearranging gives $x^{2}+y^{2}+$ $2 a x+b=0$.

Furthermore, $C_{1}$ can be rewritten as $(x+a)^{2}=y-\left(b-a^{2}\right)$, and the vertex of $C_{1}$, $\left(-a, b-a^{2}\right)$, lies inside circle $C_{2}$. Therefore, $(-a)^{2}+\left(b-a^{2}\right)^{2}+2 a(-a)+b<0$, which simplifies to $\left(b-a^{2}\right)\left(b-a^{2}+1\right)<0$, solving this gives $a^{2}-1<b<$ $a^{2}$.
Hence, when $a, b$ satisfy $a^{2}-1<b<a^{2}$, the vertex of $C_{1}$ is always inside the circle $C_{2}$ with diameter $A B$.