Prove that two circles cannot intersect at more than two points

Text Solution

Solution : To prove: Two distinct circles cannot intersect each other in more than two points.<br> Proof: Suppose that two distinct circles intersect each other in more than two points.<br>So, These points are non-collinear points.<br> Three non-collinear points determine one and only one circle.<br> Since there should be only one circle. Therefore, from those three points, 2 circles cannot pass.<br> This contradicts the given, which shows that our assumption is wrong.<br> Hence, two distinct circles cannot intersect each other in more than two points.

Suppose two circles intersect in three points A,B,C,
Then A,B,C are non-collinear. So, a unique circle passes through these three points. This is contradiction to the face that two given circles are passing through A,B,C. Hence, two circles cannot intersect each other at more than two points.

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:

$$(0:\pm\sqrt{r^2-4^2}:1) \quad\text{ and }\quad (1:\pm i:0)$$