Which system has exactly one real solution?
{y=x2y=x+2\begin{cases} y = x^2 \\ y = x + 2 \end{cases}{y=x2y=x+2
{y=x2−2x+1y=0\begin{cases} y = x^2 - 2x + 1 \\ y = 0 \end{cases}{y=x2−2x+1y=0
{y=x2y=−2\begin{cases} y = x^2 \\ y = -2 \end{cases}{y=x2y=−2
{y=(x−1)2+1y=3\begin{cases} y = (x-1)^2 + 1 \\ y = 3 \end{cases}{y=(x−1)2+1y=3