-6. The parabola $y=x^{2}+23 x+1$ intersects the line $y=2 a x+2 a b$ at most one point. Then the maximum value of $a^{2}+b^{2}$ is ).
(A) 1
(B) $\frac{\sqrt{3}}{2}$
(C) $\frac{\sqrt{2}}{2}$
(D) 0
6. (A).

From $\left\{\begin{array}{l}y=x^{2}+2 b x+1, \\ y=2 a x-2 a b\end{array}\right.$ we get
$$
x^{2}+2(b-a) x+1-2 a b=0 .
$$

Since the parabola and the line have at most one intersection point,
$$
\therefore \Delta=4(b-a)^{2}-4(1-2 a b) \leqslant 0 \text {, }
$$

which means $a^{2}+b^{2} \leqslant 1$.