8. Find all integer values of the parameter \(a\) for which the system has at least one solution

\[
\left\{\begin{array}{l}
y-2=x(x+2) \\
x^{2}+a^{2}+2 x=y(2 a-y)
\end{array}\right.
\]

In the answer, specify the sum of the found values of the parameter \(a\).
Solution. Transform the system

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

Consider the second equation of the system

$$
y^{2}-y(2 a-1)+a^{2}-2=0, \quad D=(2 a-1)^{2}-4\left(a^{2}-2\right)=9-4 a
$$

For the existence of a solution, the following conditions must be satisfied

$$
\left\{\begin{array} { l } 
{ D = 9 - 4 a \geq 0 } \\
{ [ \begin{array} { l } 
{ f ( 1 ) = a ^ { 2 } - 2 a > 0 } \\
{ \frac { 2 a - 1 } { 2 } > 1 } \\
{ f ( 1 ) = a ^ { 2 } - 2 a \leq 0 }
\end{array} }
\end{array} \Rightarrow \left\{\begin{array}{l}
a \leq 9 / 4 \\
{\left[\begin{array}{l}
\left\{\begin{array}{l}
a-2)>0 \\
2 a-1>2
\end{array}\right. \\
a(a-2) \leq 0
\end{array}\right.}
\end{array} \Rightarrow a \in[0,9 / 4]\right.\right.
$$

Summing the integer values of the parameter, we get $0+1+2=3$.

Answer: 3.