10.3. Find all values of the parameter $a$ for which the equation $x^{2}+y^{2}+2 x=|x-a|-1$ has two solutions.
Answer: such $a$ does not exist. Solution. We have one equation with two unknowns $x$, y. In the general (non-exceptional) case, it represents some curve on the coordinate plane, i.e., the equation has an infinite set of solutions. Let's check that the given equation is not an exception. Write the equation in the form $(x+1)^{2}+y^{2}=|x-a|$. For $x=-1$, there are already two solutions (for any $a \neq-1$), corresponding to two values of $y= \pm \sqrt{a+1}$. For $a \neq-1$ and for values of $x$ close to -1, the number $|x-a|-(x+1)^{2}$ will be positive, and thus, for each such $x$ there are two values of $y$ (and thus, the equation has an infinite set of solutions). It remains to consider the case when $a=-1$. In this case, we have the equation $(x+1)^{2}+y^{2}=|x+1|$. Note that the square of a number is less than the modulus of this number if the number is less than one in modulus, so the quantity $|x+1|-(x+1)^{2}$ will be positive when $|x+1|<1$. Therefore, even when $a=-1$, there will be an infinite set of solutions.