Solution
= Solution
Substituting $y=1/2$ into the <absolute value> relation gives $1/2=1-x^2$, so $x^2=1/2$. All real intersections are
$$
\boxed{(x,y)=\left(\frac1{\sqrt2},\frac12\right),\quad
\left(-\frac1{\sqrt2},\frac12\right).}
$$
= Solution
Substituting $y=1/2$ into the <absolute value> relation gives $1/2=1-x^2$, so $x^2=1/2$. All real intersections are
$$
\boxed{(x,y)=\left(\frac1{\sqrt2},\frac12\right),\quad
\left(-\frac1{\sqrt2},\frac12\right).}
$$