Answer
The intersection points of the graphs are $(0,0)$ and $(3,\pm \sqrt{3})$.
Work Step by Step
From the graph, the equations $x^2+y^2=4x$ and $x=y^2$ intersect about the points $(0,0)$ and $(3,\pm 1.7)$.
Let us check.
Substitute $x=y^2$ into the first equation to solve for $y$:
$(y^2)^2+y^2=4y^2$
$y^4-3y^2=0$
$y^2(y^2-3)=0$
So,
$y^2=0\to y=0$
$y^2-3=0\to y=\pm \sqrt{3}\approx \pm 1.73$
