10,11
A system of second-order equations
$x^{2}-y^{2}=0$

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

generally has four solutions. For what values of $a$ does the number of solutions of the system decrease to three

or two?
From the first equation, we get $y= \pm x$. Substituting this expression into the second equation, we obtain $(x-a)^{2}+x^{2}=$ 1. $\quad(1)$

The number of solutions to the system decreases to three if one of the solutions to equation (1) becomes zero.

Substituting $x=0$ into (1), we get $a^{2}=1$, that is, $a= \pm 1$.

The number of solutions to the system decreases to two if equation (1) has a single root (i.e., two coincident roots). Setting the discriminant of equation (1) to zero, we get $a= \pm \sqrt{2}$.

## Answer

The number of solutions decreases to three when $a= \pm 1$ and to two when $a= \pm \sqrt{2}$.