Task 6. (15 points) On the plane xOy, indicate all points through which no curve passes, given by the equation

$$
\mathrm{a} x^{2}+(1-6 a) x+2-a-2 y+a y^{2}=0
$$
# Solution:

Let's represent the equation as linear with respect to the parameter a.

$\mathrm{a}\left(x^{2}-6 x-1+y^{2}\right)=-x-2+2 y$. If this equation does not have solutions for any a, we will find the points (x; y) through which none of the curves pass. $\left\{\begin{array}{c}x^{2}-6 x-1+y^{2}=0 \\ -x-2+2 y \neq 0\end{array}\right.$ or $\left\{\begin{array}{c}10=y^{2}+(x-3)^{2} \\ y \neq \frac{x+2}{2}\end{array}\right.$, we obtain the circle

$10=y^{2}+(x-3)^{2}$ without the points $(0 ; 1)$ and $(4 ; 3)$.

Answer: $10=y^{2}+(x-3)^{2}$ without the points $(0 ; 1)$ and $(4 ; 3)$.