247. System of two quadratic equations

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

generally has four solutions. For what values of \( a \) does the number of solutions of this system decrease to three or to two?
247. From the first equation of the system, we immediately obtain:

$$
y^{2}=x^{2}, y= \pm x
$$

Substituting this value of $y^{2}$ into the second equation, we find:

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

- a quadratic equation which, generally speaking, gives two values of $x$. Since each value of $x$ corresponds to two values of $y$, the system has four solutions in total.

The number of solutions of the system decreases to three if one of the values of $x$ is zero; the zero value of $x$ (and only the zero value) corresponds to a single value of $y=0$, not two different values of $y= \pm x$. Substituting $x=0$ into equation (*), we get:

$$
a^{2}=1, \quad a= \pm 1
$$

only at these values of $a$ does the system have three solutions.

The number of solutions of the system decreases to two if the equation for $x$ has a single value. For the quadratic equation $(x-a)^{2}+x^{2}=1$, or $2 x^{2}-2 a x+a^{2}-1=0$, to have a single root (two coinciding values of the root), it is necessary that:

$$
a^{2}-2\left(a^{2}-1\right)=0, a^{2}=2, a= \pm \sqrt{2}
$$

at these values of $a$ the system has two solutions.