1. Find real numbers $a \neq 0, b \neq 0$, such that the graphs of the functions $y=a x^{2}$ and $y=a x+b$ intersect at two distinct points, and the coordinates of the intersection points satisfy the equation
$$
x^{2}-4 a x+y^{2}-2 b y=0 \text {. }
$$
1. Let $a, b$ be the desired values.

Then there exist $\left(x_{1}, y_{1}\right), \left(x_{2}, y_{2}\right)$ such that
$$
y_{k}=a x_{k}^{2}=a x_{k}+b \quad (k=1,2), \quad x_{1} \neq x_{2},
$$

and for $k=1,2$ we have
$$
\begin{aligned}
0 & =x_{k}^{2}-4 a x_{k}+y_{k}^{2}-2 b y_{k} \\
& =x_{k}^{2}-4 a x_{k}+\left(y_{k}-b\right)^{2}-b^{2} \\
& =x_{k}^{2}-4 a x_{k}+a^{2} x_{k}^{2}-b^{2} \\
& =\left(a+\frac{1}{a}\right)\left(a x_{k}+b\right)-4 a x_{k}-b^{2} \\
& =\left(a^{2}-4 a+1\right) x_{k}+a b+\frac{b}{a}-b^{2} .
\end{aligned}
$$

Thus, $a^{2}-4 a+1=0 ; b=a+\frac{1}{a}$.
Therefore, $a=2 \pm \sqrt{3}, b=4$.
Conversely, it is easy to verify that $a=2 \pm \sqrt{3}, b=4$ satisfy the problem's requirements.