11.2. Find all values of the parameter $a \in R$, for which the system

$$
\left\{\begin{array}{c}
y=\sqrt{5+4 x-x^{2}}+2 \\
y=\sqrt{9-a^{2}+2 a x-x^{2}}+a
\end{array}\right.
$$

has exactly one solution in $R \times R$.
Solution. The initial system is equivalent to the following system:

$$
\left\{\begin{array} { c } 
{ ( y - 2 ) ^ { 2 } = 5 + 4 x - x ^ { 2 } } \\
{ y - 2 \geq 0 , \quad 5 + 4 x - x ^ { 2 } \geq 0 , } \\
{ ( y - a ) ^ { 2 } = 9 - a ^ { 2 } + 2 a x - x ^ { 2 } , } \\
{ y - a \geq 0 , \quad 9 - a ^ { 2 } + 2 a x - x ^ { 2 } \geq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
(y-2)^{2}+(x-2)^{2}=9 \\
y-2 \geq 0, \quad(x-2)^{2} \leq 9 \\
(y-a)^{2}+(x-a)^{2}=9 \\
y-a \geq 0, \quad(x-a)^{2} \leq 9
\end{array}\right.\right.
$$

The geometric figure determined by the relations written in the first two lines of the system is the semicircle $T_{2}$ with center $(2 ; 2)$ and radius 3 (in addition, $|x-2| \leq 3 \Rightarrow x \in[-1 ; 5]$). Similarly, the geometric figure determined by the relations written in the third and fourth lines of the system is the semicircle $T_{a}$ with center at the point $(a ; a)$ and radius 3 (in addition, $x \in[a-3 ; a+3]$).

![](https://cdn.mathpix.com/cropped/2024_06_06_4435721db0f84129ef86g-2.jpg?height=931&width=1542&top_left_y=682&top_left_x=157)

Figure 11.2

Since the radii of these semicircles $T_{a}$ are the same (and equal to 3), and their centers $(a ; a)$ lie on the line $y=x$, then (see Figure 11.2) these semicircles $T_{a}$ can intersect the fixed semicircle $T_{2}$ only when $-1 \leq a \leq 5$. For $a \neq 2$, these semicircles $T_{a}$ intersect $T_{2}$ at one point, and for $a=2$ we have $T_{2} \equiv T_{a}$. Therefore, $a \in[-1 ; 5] \backslash\{2\}$.