3. Find all values of the real parameter $m$ for which the system of equations

$$
\begin{aligned}
x^{2}+y^{2} & =4 \\
(x+m)^{2}+(y-m)^{2} & =1
\end{aligned}
$$

has exactly one solution.
3. First solution. The second equation can be rewritten as $x^{2}+2 m x+m^{2}+y^{2}-2 m y+m^{2}=1$, and using the first equation here, we get $4+2 m(x-y)+2 m^{2}=1$, i.e., $2 m(x-y)=-2 m^{2}-3$. We immediately see that for $m=0$ there is no solution. Now assume $m \neq 0$. Then we get $x=y-\frac{2 m^{2}+3}{2 m}$. Substituting this into the first equation of the system, we obtain $\left(y-\frac{2 m^{2}+3}{2 m}\right)^{2}+y^{2}=4$, i.e.,

$$
2 y^{2}-\frac{2 m^{2}+3}{m} y+\frac{4 m^{4}-4 m^{2}+9}{4 m^{2}}=0
$$

This equation has exactly one solution if and only if its discriminant is 0, that is:

$$
\begin{aligned}
0 & =\frac{\left(2 m^{2}+3\right)^{2}}{m^{2}}-8 \cdot \frac{4 m^{4}-4 m^{2}+9}{4 m^{2}} \\
& =\frac{4 m^{4}+12 m^{2}+9-8 m^{4}+8 m^{2}-18}{m^{2}}=-\frac{4 m^{4}-20 m^{2}+9}{m^{2}}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_04_f254ba4f25ce7f8d9df1g-45.jpg?height=454&width=507&top_left_y=1633&top_left_x=189)

Ok 2017 3B 3
Thus, with the substitution $t=m^{2}$, we have $4 t^{2}-20 t+9=0$, whose solutions are $t_{1 / 2}=\frac{20 \pm \sqrt{400-144}}{8}=\frac{20 \pm 16}{8}$, i.e., $t_{1}=\frac{9}{2}$ and $t_{2}=\frac{1}{2}$. Therefore, there are four possible values for the parameter $m: m= \pm \frac{3}{\sqrt{2}}$ and $m= \pm \frac{1}{\sqrt{2}}$.

Second solution. In the coordinate plane, consider the points $O(0,0), A(x, y)$, and $M(-m, m)$. Then $O A^{2}=x^{2}+y^{2}$ and $M A^{2}=(x+m)^{2}+(y-m)^{2}$, so from the equality in the problem, it follows that $A$ is a point for which $O A=1$ and $M A=2$. In other words, for a given value of $m$, the point $A$ is at the intersection of the circles $k_{1}(O, 1)$ and $k_{2}(M, 2)$.

The solution $(x, y)$ of the given system is unique if and only if the circles $k_{1}$ and $k_{2}$ are tangent, which is equivalent to the condition $O M=2 \pm 1$, i.e., $O M=3$ or $O M=1$. Given that $O M=|m| \sqrt{2}$, it follows that $m= \pm \frac{3}{\sqrt{2}}$ or $m= \pm \frac{1}{\sqrt{2}}$.