Answer: the system has no solution.
Step-by-step explanation:
\displaystyle\\
\left \{ {{x^2y=16\ \ \ \ \ (1)} \atop {x^2+4y+16=0\ \ \ \ \ (2)}} \right. .\\
Multiply\ both\ sides\ of\ the\ equation\ (2)\ by\ y\ (y\neq 0):\\
x^2y+4y^2+16y=0\\
We\ substitute\ equation\ (1)\ into\ equation\ (2):\\
16+4y^2+16y=0\\
4y^2+16y+16=0\\
4*(y^2+4y+4)=0\\
4*(y^2+2*y*2+2^2)=0\\
4*(y+2)^2=0\\
Divide\ both\ sides\ of\ the \ equation\ by\ 4:\\
(y+2)^2=0\\
(y+2)*(y+2)=0\\
So,\ y+2=0\\
y=-2.\\

So when you solve, it’s going to be y=60x-30. So on the graph, you start at -30 and add 60 up and one box over.
to find the area you have to do:
Area = π( D/2)^2 and in this case D = 19 so..
π(19/2)^2
=90.25π
=283.52873698648
Answer: 283.53