Extending on Mockingbird's excellent answer, it's nice and instructive to calculate the 4 intersection points of two concentric circles of different radius.
Without loss of generality, let the circles be centered at the origin, so that their defining equations are
$$x^2+y^2 = r_k^2 \tag 1$$
where $r_1\neq r_2\in \Bbb R^+$ are the radii. A circle is a curve of order 2, and for Bézout's Theorem to yield $2\cdot 2=4$ intersections between 2 circles, we need to switch to projective space. This turns the affine definitions (1) into their projective counter-parts$^4$
$$X^2+Y^2 = r_k^2 Z^2 \tag 2$$
where any two points $(X:Y:Z)$ and $(\lambda X:\lambda Y:\lambda Z)$ for some $\lambda\neq0$ are regarded as being identical$^1$.
One more ingredient to Bézout's Theorem is needed, namely that the field in which the coordinates live must be algebraically closed$^2$, so we take the Complex Numbers and $X, Y, Z\in\Bbb C$.
In order to determine conditions for the intersections of the two circles, rewrite (2) as
$$0 = X^2+Y^2 - r_1^2 Z^2 = X^2+Y^2 - r_2^2 Z^2 \tag 3$$
which implies $r_1^2Z^2=r_2^2Z^2$, which implies $Z^2=0$ as $r_1\neq r_2$. It follows that $X$ and $Y$ must satisfy $0 = X^2+Y^2$ and thus
$X^2 = -Y^2$
which has solutions
$$X = \pm iY \tag 4$$
where $i=\sqrt{-1}$. So we get the two solutions$^5$
$$(\pm iY:Y:0) = (\pm i:1:0) \tag 5$$
because we must have$^3$ $Y\neq0$, and we can divide through by $Y$.
So we found two intersections; but where are the other two intersections promised by Bézout's Theorem? We only get 4 intersections when taking into account multiplicities of the solutions, which is 2 for both points, because $Z=0$ is a double zero of $Z^2=0$.
So in this example we needed all the fancy features to get 4 solutions: Algebraic closedness of the field, projective space, and multipicity.
Footnotes:
$^1$Notice that in the case $Z \neq 0$ we can take $\lambda=1/Z$ so that $(X:Y:Z)=(x:y:1)$ with $x=\lambda X$ and $y=\lambda Y$. The projective point $(x:y:1)$ can then be identified with the point $(x,y)$ in the "ordinary" plane. The remaining points of the form $(X:Y:0)$ are sometimes called "points at infinity", and they are not part of the ordinary plane.
$^2$Which means that each polynomial with coefficients over the field $K$ must have a root in $K$. For example, $\Bbb R$ is not algebraically closed because the polynomial $x^2+1$ has no root in $\Bbb R$.
$^3$The point $(0:0:0)$ is not an element of projective space.
$^4$The procedure that can be used is: Substitute $x=X/Z$ and $y=Y/Z$, and then multiply through by $Z^2$ in order to clear denominators.
$^5$Notice that the representation of the solutions is by no means unique, for example $(i:1:0) = (-1:i:0) = (1:-i:0) = (2:-2i:0)$ etc.
Addendum:
As an additional treat, let's investigate what happens if two circles do have intersections in the plane. To that end, let's take the circles of radius $5$ around $(4,0)$ and $(-4,0)$. The defining affine equations are $(x\pm4)^2+y^2=5^2$, which become
$$(X\pm4 Z)^2 + Y^2 = 5^2Z^2$$
in projective space. They yield the condition
$$0 = (X-4Z)^2 + Y^2 -5^2Z^2 ~=~ (X+4Z)^2 + Y^2 - 5^2Z^2\tag 7$$
It implies $(X-4Z)^2 = (X+4Z)^2$, which has the solution $XZ=0$. So there are three cases, and when plugging them back into $(7)$ we get:
$$\begin{align}
X=0, Z=1 &\quad\implies\quad 0 = 4^2 + Y^2 -5^2\tag {7.1} \\
X=1, Z=0 &\quad\implies\quad 0 = 1^2 + Y^2\tag {7.2} \\
X=0, Z=0 &\quad\implies\quad 0 = Y^2\tag {7.3} \\
\end{align}$$
The first case has two solutions $Y=\pm3$. The second case has two solutions $Y=\pm i$. The 3rd case is not a solution because $(0:0:0)$ is not a projective point. Hence the 4 solutions are:
$$(0:\pm 3: 1);\quad (1:\pm i:0)$$
The first 2 solutions are the affine ones you also get in the Euclidean plane as $(x,y) = (0,\pm3)$. And the latter two are solutions at "infinity".
It's also easy to see now what happens when these two off-center circles do not intersect in $\Bbb R^2$: $X$ and $Z$ are not affected when we change the radius from $5$ to $r$. And in the case $Z=1$ we get $Y=\pm\sqrt{r^2-4^2}$ which is imaginary when $r<4$ (and a 2-fold solution in the case $r=4$ of kissing circles). The solutions are then:
$$(0:\pm\sqrt{r^2-4^2}:1) \quad\text{ and }\quad (1:\pm i:0)$$