7. In the Cartesian coordinate system, it is known that the parabola $y=k x^{2}(k>0)$ intersects the circle $(x-a)^{2}+(y-b)^{2}=r^{2}$ at least at 3 points, one of which is the origin, and the other two lie on the line $y=k x+b$. What is the minimum value of the real number $b$? $\qquad$ .
According to the problem, the circle $(x-a)^{2}+(y-b)^{2}=r^{2}$ passes through the origin $\Rightarrow x^{2}+y^{2}-2 a x-2 b y=0$.
Solving the system $\left\{\begin{array}{l}y=k x+b, \\ y=k x^{2}\end{array} \Rightarrow k x^{2}-k x-b=0\right.$,
Solving the system $\left\{\begin{array}{l}y=k x+b, \\ x^{2}+y^{2}-2 a x-2 b y=0\end{array} \Rightarrow\left(k^{2}+1\right) x^{2}-2 a x-b^{2}=0\right.$,
Since the roots of (1) and (2) are the same $\Rightarrow b=\frac{k^{2}+1}{k} \geqslant 2$, equality holds when $k=1$.