3-ча 3. A system of second-order equations

$$
\left\{\begin{aligned}
x^{2}-y^{2} & =0 \\
(x-a)^{2}+y^{2} & =1
\end{aligned}\right.
$$

generally has four solutions. For what values of $a$ does the number of solutions of the system decrease to three or to two?
Solve 3. Answer: the number of solutions decreases to three when $a= \pm 1$, and the number of solutions decreases to two when $a= \pm \sqrt{2}$.

From the first equation, we get $y= \pm x$. Substituting this expression into the second equation, we obtain

$$
(x-a)^{2}+x^{2}=1
$$

The number of solutions of the system decreases to three if one of the solutions of equation (1) becomes zero. Substituting $x=0$ into (1), we get $a^{2}=1$, i.e., $a= \pm 1$. The number of solutions of the system decreases to two if equation (1) has a unique root (i.e., two coincident roots). Setting the discriminant of equation (1) to zero, we get $a= \pm \sqrt{2}$.