Answer
The solution is $\left\{ \left( 0,0 \right),\left( -2,2 \right),\left( 2,2 \right) \right\}$
Work Step by Step
The equations are,
$\begin{align}
& {{x}^{2}}+{{\left( y-2 \right)}^{2}}=4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\
& {{x}^{2}}-2y=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \\
\end{align}$
From equation (II):
${{x}^{2}}=2y $ (III)
Substitute ${{x}^{2}}=2y $ in equation (I):
$\begin{align}
& 2y+{{\left( y-2 \right)}^{2}}=4 \\
& 2y+{{y}^{2}}-4y+4=4 \\
& {{y}^{2}}-2y=0 \\
& y\left( y-2 \right)=0 \\
& y=0,2
\end{align}$
Substitute $ y=0$ in equation (III):
$\begin{align}
& {{x}^{2}}=2\left( 0 \right) \\
& {{x}^{2}}=0 \\
& x=0
\end{align}$
Substitute $ y=2$ in equation (III):
$\begin{align}
& {{x}^{2}}=2\left( 2 \right) \\
& {{x}^{2}}=4 \\
& x=\pm 2
\end{align}$
Thus, the solution is $\left\{ \left( 0,0 \right),\left( -2,2 \right),\left( 2,2 \right) \right\}$.
