11.7. Find all values of the real parameter \( a \) for which the system

\[
\left\{\begin{array}{c}
x^{2}+5 x+y^{2}-y-|x-5 y+5|=51 \\
a x-y-5 a+2=0
\end{array}\right.
\]

has exactly 2 distinct real solutions.
Solution. We examine the expression $E(x, y)=x-5 y+5$ which is in the modulus. We have two possible cases.

Case 1. Let $E(x, y) \geq 0$, i.e., $y \leq \frac{x}{5}+1$. We obtain that the initial system is equivalent to the system of equations

$$
\left\{\begin{array} { c } 
{ x ^ { 2 } + 5 x + y ^ { 2 } - y - x + 5 y - 5 = 5 1 } \\
{ y - 2 = a ( x - 5 ) }
\end{array} \text {, which can be written in the equivalent form } \left\{\begin{array}{c}
(x+2)^{2}+(y+2)^{2}=8^{2} \\
y=a(x-5)+2
\end{array}\right.\right.
$$

![](https://cdn.mathpix.com/cropped/2024_06_06_888e1c20cbcb21814b01g-2.jpg?height=751&width=1042&top_left_y=296&top_left_x=39)

The equation $y=a(x-5)+2$ determines a bundle of lines passing through the point $A(5 ; 2)$. One of these lines is the boundary line $y=\frac{x}{5}+1$ (for $a=\frac{1}{5}$), which delimits the half-plane determined by the condition $E(x, y) \geq 0$.

The second boundary line is the line $y=a_{1}(x-5)+2$, which is tangent to the circle $(x+2)^{2}+(y+2)^{2}=8^{2}$. From these considerations, the equation $(x+2)^{2}+\left(a_{1}(x-5)+4\right)^{2}=8^{2}$ has exactly one solution. This equation can be equivalently written as

$$
\left(1+a_{1}^{2}\right) x^{2}-\left(10 a_{1}^{2}-8 a_{1}-4\right) x+\left(25 a_{1}^{2}-40 a_{1}-44\right)=0
$$

and, since this equation has exactly one solution, it follows that $0=\Delta=60 a_{1}^{2}+224 a_{1}+192$. We obtain two possible values of the parameter $a_{1}:-\frac{4}{3}$ and $-2.4$ (for the two tangents drawn from the point $A$ to the given circle), but we choose the value $a_{1}=-2.4$, for which the imposed condition $E(x, y) \geq 0$ is verified (as can be seen from the graph).

Thus, in this case, we obtain:

1) if $a \in(-\infty ;-2.4)$, then we have two points of intersection, i.e., the system has two solutions;
2) if $a=-2.4$, then we have one point of intersection, i.e., the system has only one solution;
3) if $a \in(-2.4 ; 0.2)$, then either we have no points of intersection, or these points are above the boundary line $y=\frac{x}{5}+1$, i.e., the system has no solutions;
4) if $a \in[0.2 ;+\infty)$, then we have two points of intersection, i.e., the system has two solutions.

Case 2. Let $E(x, y) < \frac{x}{5}+1$. We obtain that the initial system is equivalent to the system of equations

$\left\{\begin{array}{c}x^{2}+5 x+y^{2}-y+x-5 y+5=51 \\ y-2=a(x-5)\end{array}\right.$, which can be written in the equivalent form $\left\{\begin{array}{c}(x+3)^{2}+(y-3)^{2}=64 \\ y=a(x-5)+2\end{array}\right.$.

![](https://cdn.mathpix.com/cropped/2024_06_06_888e1c20cbcb21814b01g-2.jpg?height=743&width=1154&top_left_y=2067&top_left_x=37)

The reasoning is analogous to the previous case, only the second boundary line (which is tangent to the circle $\left.(x+3)^{2}+(y-3)^{2}=64\right)$, is the line $x=5$. Thus, in this case, we obtain:

1) if $a \in(-\infty ; 0.2)$, then we have two points of intersection, i.e., the system has two solutions;
2) if $a \in[0.2 ;+\infty)$, then either we have no points of intersection, or these points are below the boundary line $y=\frac{x}{5}+1$, i.e., the system has no solutions.

Combining the two cases, we obtain:

1) if $a \in(-\infty ;-2.4)$, then the system has 4 solutions;
2) if $a=-2.4$, then the system has 3 solutions;
3) if $a \in(-2.4 ; 0.2)$, then the system has 2 solutions;
4) if $a \in[0.2 ;+\infty)$, then the system has 2 solutions.

Thus, the system has exactly 2 solutions if and only if $a \in(-2.4 ;+\infty)$.