I have been trying to solve this problem for a week now. It goes like this: Find all values of $a$ for which the system $$ \begin{cases} x^2-2x+y^2 = 1 \\[1ex] \dfrac{x+|x|}{y-a}=2 \end{cases} $$ has only one solution for $x$ and $y$.
I found one such value $a=-3$, however the answer sheet says $a\in(-1,1)\cup\{-3\}$.
Note that $$x^2-2x+y^2=1\iff (x-1)^2+y^2=2$$ which is the circle whose center is $(1,0)$ with the radius $\sqrt 2$.
Also, if $x\le 0$, then $x+|x|=x-x=0$. So, we need $x\gt 0$ and $$\begin{align}\frac{x+|x|}{y-a}=2\ \text{and}\ x\gt 0&\iff x+x=2(y-a)\ \text{and}\ y-a\not=0\ \text{and}\ x\gt 0\\&\iff y=x+a\ \text{and}\ x\gt 0.\end{align}$$
Hence, consider the case when the circle and the half line $y=x+a\ (x\gt 0)$ have only one intersection.
P.S. Since you haven't studied the equation of circle, then you can do as the following : Consider the case when there is only one positive root in $$(x-1)^2+(x+a)^2=2\iff f(x)=2x^2+2(-1+a)x+a^2-1=0\tag1$$
Case 1 : $D/4=(-1+a)^2-2(a^2-1)=0\iff (a+3)(a-1)=0\iff a=-3,1$. If $a=-3$, then $(1)\iff (x-2)^2=0$ has only one positive root. If $a=1$, then $(1)\iff x^2=0$ has $x=0$.
Case 2 : $D/4\gt 0\iff -3\lt a\lt 1$. The condition we need is $f(0)\lt 0\iff -1\lt a\lt 1$ (Why?).
Hence, the answer is what the answer sheet says.