Given a point $p = (x,y) \in \mathbb{R}^2$, define its lifting $l(p)$ as the point $l(p) = (x,y, x^2+y^2) \in \mathbb{R}^3$. Then as per the image below, I wish to show that the lifted circle $l(C)$ is contained in a unique plane $h_C \subset \mathbb{R}^3$.
Given a circle centered at $(0,0)$ (I'll change my argument for a general center later) with radius $r$, then points in the image $l(C)$ are of the form $(r \cos \theta, r \sin \theta, r^2)$ for $\theta \in [0, 2 \pi]$.
What is the normal for such a plane?
The point-normal form of the equation of a plane basically says that a plane is the locus of points $\mathbf p$ for which, for some fixed $\mathbf n\ne0$, the quantity $\mathbf n\cdot\mathbf p$ is constant. The vector $\mathbf n$ is the plane’s normal: it’s perpendicular to the plane, so for an arbitrary circle on the $x$-$y$ plane with center $(h,k)$ and radius $r$, take three noncolinear points on the circle, lift them and then use the usual subtract-and-cross-product computation to get a normal to the plane defined by these three points.
For instance, we can take the points $(h+r,k,(h+r)^2+k^2)$, $(h,k+r,h^2+(k+r)^2)$ and $(h,k-r,h^2+(k-r)^2)$. The resulting normal, after taking out some extraneous common factors, is $\mathbf n = (2h,2k,-1)$. (This is also, not coincidentally, the normal to the graph of $z=x^2+y^2$ at the center of the circle.) So, for an arbitrary point $\mathbf p = (x,y,x^2+y^2)$ on this circle, $\mathbf n\cdot\mathbf p$ must depend only on $h$, $k$ and $r$ in order for the image of the circle to be planar. You can perform this calculation with either a parameterization $$\mathbf p = \left(h+r\cos t,k+r\sin t,(h+r\cos t)^2+(k+r\sin t)^2\right)\tag1$$ of the circle or use the standard Cartesian equation of the circle to obtain $$\mathbf p = \left(x,y,2(hx+ky)-(h^2+k^2-r^2)\right).\tag2$$
An equivalent approach is to forgo computing a normal vector explicitly and instead evaluate $$\begin{vmatrix}h+r&k&(h+r)^2+k^2&1 \\ h-r&k&(h-r)^2+k^2&1\\h&k+r&h^2+(k+r)^2&1\\h+r\cos t&k+r\sin t&(h+r\cos t)^2+(k+r\sin t)^2&1\end{vmatrix}.$$ Since we know the points represented by the first three rows are not colinear (verify this!), this determinant vanishes iff the last point is coplanar with the first three. Instead of expanding this determinant directly, one can row-reduce the matrix to obtain $$\begin{vmatrix} r&0&2hr&0\\0&r&2kr&0\\0&0&r^2-h^2-k^2&1\\0&0&r^2-h^2-k^2&1 \end{vmatrix}.$$ The last two rows are identical, so this determinant vanishes for all $t$. You could also use the expression in (2) for the last row of the matrix. Most of the computations to reduce it will be the same.