Not sure if this is a misconception of yours or not, but I would first like to address this regardless: a circle is not a quadratic. Although there is a degree two x variable, there's also a degree two y variable in the general equation which prevents us from classifying a circle as a quadratic. You may then ask: what type of equation is a circle? Well, for the purposes of the syllabus, a circle is just a circle. That's all there is to it (If you're interested, you could search up "conics" which can give you more insight into a circle, but that's beyond the Methods curriculum, so don't stress if you don't understand it all). Also apologies if you already know that a circle is not a quadratic.
What you did was find the midpoint of the two x-intercepts, (4,0) and (-4,0), of the circle, getting (0,0) and assuming that this is the center of the circle. The issue with doing this is that, using your x-intercepts, you can only deduce the x-coordinate of the center, which is 0, in this case. However, we cannot assume that the center has the same y-coordinate as these x-intercepts (i.e. we cannot assume the y-coordinate is zero). Using your method, we would actually have to find the midpoint of the y-intercepts, which are (0,8) and (0,-2), and take the y-coordinate (3) of the midpoint (which is (0,3) ) to get the y-coordinate of the circle's center. We would then put these pieces of information together to get the center of (0,3).
Really nice help, p0kem0n21! But I wanted to jump in on this line right here:
From this, you can see that the method is a bit lengthy and requires a bunch of substitution and quadratic solving.
There's actually a very easy thing you can do here once you know a point on the circle and the centre - after all, the radius is just the distance from the centre to a point on the circle. So, if I know the centre of the circle is (0,3), and that one of the x-intercepts is (4,0), then the distance from the centre of the circle to the x-intercept is:
\[
D=\sqrt{(4-0)^2+(0-3)^2}=\sqrt{16+9}=\sqrt{25}=5
\]
So, if the centre is (0,3), the radius is 5, and the equation of the circle is \(x^2+(y-3)^2=5^2\)