Why can't two circles intersect at more than 2 points? According to Bézout's Theorem, two polynomials of degree $m,n$ intersect at most at $mn$ points. So, two circles should intersect at most at $4$ points as well. But I have so far known that 2 circles intersect at most at $2$ points. Why?
Why doesn't Bézout's theorem apply for circles? Is this a special case of the theorem?
Can anyone explain this in easy terms? (I'm only in high school!)
 A: According to your statement of Bézout's Theorem, (the curves defined by) two polynomials of degree $m,n$ intersect at most at $mn$ points.
Since $2 \le 4$ everything's fine.
There is however a stronger version of Bézout's Theorem: the curves defined by two polynomials of degree $m,n$ intersect at exactly $mn$ points.
To get this, we need to add some technicalities here and there:

*

*you require that the polynomials are distinct and irreducible (they can't be split in simpler factors). This assures that the curves do not have big pieces in common: we want they just intersect at points.

*you don't use the real numbers, but a bigger number set (the complex numbers, if you know what they are) to be sure that there are indeed roots.
In the case of two circles in the real plane, they may not intersect.

*You count roots with multiplicities: each root may be 'hit' more than once, as in the case of tangent circles.

*you don't draw your curves in a plane, but in a bit more sophisticated geometric environment, known as projective plane, in which two curves may encounter at special points (points at infinity).

The reason why you always get at most $2$ is that the other two solutions do indeed appear when you consider your curves as lying in this special environment: the complex projective plane.
A: One answer has to do with con-formal mapping. Basically you map the points in a plane in such a way that angles between tangents to curves at their point of intersection are preserved, except perhaps at isolated points. The number of points of intersection apart from these isolated cases is also preserved. So for instance, if you can conformally map two circles into a circle and a line, then you have at most two points of intersection between the circle and the line; so you can have no more than two between the original circles too.
It turns out that two circles can be conformally mapped into a circle and a line, so there you are. Two ellipses, or one ellipse and a circle, cannot be conformally mapped in this way, so in these cases the number of points of intersection is not always reduced from four to two.
A: 
According to Bézout's Theorem, two polynomials of degree $m,n$ intersect at most at $mn$ points. So, two circles should intersect at most at $4$ points as well. But I have so far known that 2 circles intersect at most at $2$ points.

To be precise, Bézout's theorem states that if you have two different polynomials of degree $m$ and $n$ (and those polynomials have no common factor which is not a constant), then those polynomials do not intersect at more than $mn$ points. As you know, a circle can be defined by a polynomial of degree $2$, so Bézout's theorem tells you that if you have two different circles, then those circles do not intersect at more than $4$ points.

Why doesn't Bézout's theorem apply for circles?

Bézout's theorem does apply for circles. The statement "if you have two different circles, then those circles do not intersect at more than $4$ points" is completely true.
A: A circle in the plane can be described by three parameters: The centre (x, y; 2 parameters) and the radius. As a result, you can determine a circle when you are given three points on the circle. (That is of course something you will have to prove. But the centre has equal distance from any two points; and given three points A, B and C the points of equal distance from A and B for example form a line, and you have three lines which must meet at the centre).
So first you prove that given any three points A, B and C there is at most one circle with all three points on the circle. If you had two circles intersecting in three points, then both would have to be identical. So at most two intersections, unless the two circles are identical.
You might examine the question for other shapes, like ellipse, square, rectangle, etc. I think squares can intersect in eight points.
Bezoit's Theorem obviously applies. It says "not more than four intersections" which is true. It doesn't claim "Four or three intersections are always possible, no matter what restrictions there are". Actually, many pairs of circles have no intersection at all.
A: 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:
$$(0:\pm\sqrt{r^2-4^2}:1) \quad\text{ and }\quad (1:\pm i:0)$$
A: Mockingbird gave a good answer providing the general reasoning and emacs drives me nuts computed an example. However in their example the two circles do not interset in the real plane at all so I thought it would be useful to also look at an example where there are two intersections in the real plane and then find the other two.
Consider the two circles $x^2+y^2=1$ and $(x-1)^2+y^2=1$. One can compute that they intersect at the two points $x_{1/2}=\frac{1}{2}$ and $y_{1/2}=\pm \frac{\sqrt{3}}{2}$ in the real plane.
Just assuming that $x$ and $y$ are complex instead of real numbers does not yield any additional solutions.
If you homogenize the equation to move to the projective plane, the equations turn into $X^2+Y^2=Z^2$ and $(X-Z)^2+Y^2=Z^2$ where as before $[X,Y,Z]$ and $[\lambda X, \lambda Y, \lambda Z]$ are considered the same for any complex $\lambda \neq 0$.
Now substracting the two equations from each other gives $Z^2=2XZ$. This equation has two solutions, $X=\frac{1}{2}$ which yields the two solutions we already know about namely $[\frac{1}{2}, \pm \frac{\sqrt{3}}{2}, 1]$ and $Z=0$ which gives the two new solutions $[1, \pm i, 0]$.
One can even picture where these solutions are in the complex plane. If $x$ is a large real number and $y$ is $i$ times the same large real number than both equations are almost satisfied because the $1$ and $-1$ become insignificant relative to the large real number. If you let the real number go all the way to infinity the equations become satisfied exactly. The projective plane allows you to make this very handwavy argument mathematically rigorous.
A: We have two circles with different centers, which centers we align along the x-axis in the x, y plane. In particular, we place the origin of the x, y plane at the center of one circle. That circle has equation x^2 + y^2 = r^2, while the other has equation (x-h)^2 + y^2 = R^2, for some h > 0. Thus, if the circles intersect at (x, y), then we have y^2 = x^2 - r^2 = R^2 - (x-h)^2, leading to x = (r^2 - R^2 - h^2) / (2h), a unique real value of x. Hence, y^2 has a value that is either negative, zero, or positive, corresponding to the number of intersections being 0, 1, or 2 (from those being the possible numbers of y-values). QED
