My friend recently played in GeoGebra and discovered an interesting fun fact:

Given two conic sections such that they have two common tangents in two different points $A$ and $B$(conics are touching, $A$ and $B$ lie on both of them, and as such, they have common tangents). Let the intersection of those tangents be point $O$. Let us denote conics $g$ and $h$. Pick a point $F$ on $g$ and let $l$ be line tangent to $g$ in $F$ such that $l$ intersects $h$ in two points $C$ and $D$. Let $c$ and $d$ be tangents to $h$ in $C$ and $D$ respectively and let $M$ be their intersection. Prove that $O$ , $M$ and $F$ are collinear.

Of course, there are, in the Euclidean plane, degenerate cases where these points do not exist, but ignore them by assuming all intersections. We believe that this problem is easy in projective geometry, apart from the fact that there are no degenerate cases in it and talks only about incidence, but none of us had any experience, and neither did our geometry teacher succeed(we are high school students), so we ask for help.

This problem turned out to have some significance to one of our projects so we would be really grateful to see some solution soon :)

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    $\begingroup$ I added a figure (with two ellipses) to help visualise the problem. $\endgroup$ Oct 4, 2021 at 17:24
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    $\begingroup$ You're right, it's easy using projective geometry. But is this the kind of proof you want? Because there may be terms like pole/polar, cross ratio, double contact, etc that don't mean anything to you. $\endgroup$
    – brainjam
    Oct 4, 2021 at 22:25
  • $\begingroup$ @brainjam , we would like to see a solution in any form, it is easy to google unknown definitions and theorems, much easier than formulating a proof in theory you know almost nothing about 😅 $\endgroup$ Oct 5, 2021 at 16:34

3 Answers 3


enter image description here

The diagram has been redrawn. Added is the line $AB$, which is called the chord of contact (between the red and green ellipses). $X$ is the intersection of the lines $AB$ and $CD$.

What follows is a roadmap of a proof. The basic idea is that points $M,F,O$ are all on the polar of $X$ wrt the red conic $c$, and are thus collinear. We'll show this for each point, but the reader unfamiliar with projective geometry may have to consult the given references to make sense of things.

For some background on poles and polars see Wikipedia. The main facts we'll use here are (all poles and polars are wrt to $c$):

  • (i) let $p$ be the polar of $X$, and let $X',Y,Z$ be the points where a line going through $X$ is cut by $p$ and $c$. Then the cross ratio $(Y,Z;X,X')=-1$, i.e. $X'$ is the harmonic conjugate of $X$ wrt to $W,Z$. We say the line is cut harmonically by $X,X',Y,Z$.
  • (ii) if a point $P$ lies on the polar of point $Q$, then $Q$ lies on the polar of $P$.
  • (iii) if the tangents from $P$ to $c$ touch $c$ at $P',P''$, then the line $P'P''$ is the polar of $P$.

The red and green conics touch in two points $A,B$ and are said to be in double contact. One theorem we want is (see Milne, Cross Ratio Geometry, Article 266):

(*) Any tangent to [one conic] is cut harmonically at its point of contact, and at the points where it meets the chord of contact and the other conic.

The background of (*) will be discussed later, let's get to a proof of the OP.

$M$: by (iii), the line $CD$ is the polar of $M$. Since $X$ is on the polar of $M$, by (ii) $M$ is on the polar of $X$.

$O$: as for $M$, $X$ is on the polar of $O$ and therefore $O$ is on the polar of $M$.

$F$: by (*), the tangent $CD$ is cut harmonically by $C,D,F,X$. By (i), $F$ is on the polar of $X$.

So $M,O,F$ are all on the same line (the polar of $X$), and we are done.

Some remarks on theorem (*). This is a special case of a special case of Desargues' Theorem (Milne, Article 187), which requires a familiarity with projective involutions and their fixed points (which are related to systems of coaxal circles, which may be more familiar). So to understand it will require leafing backwards through the text, often guided by references to earlier articles. If a less rigorous understanding is acceptable, simply playing with the configuration using Geogebra and the its Polar and CrossRatio commands may suffice.

To see the special case, move the point $A$ in Milne's figure to coincide with $C$, and $B$ with $D$. So the general case of conics intersecting in four points becomes that of touching in two points. And in the above diagram, the tangent $CD$ is a special case of a line that intersects both conics.

  • $\begingroup$ I just posted an answer at exactly the same time as yours ! $\endgroup$
    – Jean Marie
    Oct 5, 2021 at 19:35
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    $\begingroup$ @Jean, yes, and they are different approaches. I was worried they would be the same. $\endgroup$
    – brainjam
    Oct 5, 2021 at 19:36
  • $\begingroup$ [+1] Very interesting approach indeed. $\endgroup$
    – Jean Marie
    Oct 5, 2021 at 21:38
  • $\begingroup$ Thank you for your solution, it is very clear and helped us a lot $\endgroup$ Oct 5, 2021 at 22:58

Instead of considering a general case, I am going to work on an example which - I think - is more didactic that a general case with many variables, being given the amount of new concepts and techniques (!).

Let us consider the particular case where the two conics are a circle and parabola with resp equations,

$$\begin{cases}x^2-2y+1&=&0& \ \ (parabola \ C_1)\\x^2+y^2-4y+4&=&0& \ \ (circle \ C_2)\end{cases}\tag{1}$$

enter image description here

Fig. 1 : The conic curves on which we are going to work: a circle and a parabola.

Please note that we have used the notations of the figure made by @Intelligenci Pauca with $O=(0,0)$ the origin of coordinates.

The two conic curves share 2 common contact elements (a contact element = a point + a line passing through this point), the points being $A(-1,1)$ and $B(1,1)$ and for the lines the bissector lines $x-y=0$ and $x+y=0$.

Now, circle $C_2$ being with center $(0,2)$ and radius $r=\sqrt{2}$, its generic point is

$$F=(r \cos a, \ 2+r \sin a) \ \ \text{with} \ \ r=\sqrt{2}$$

It is rather easy to determine the equation of the tangent $CD$ to $C_1$ (passing through $F$). It is:

$$\underbrace{\cos a}_u x +\underbrace{ \sin a}_v y+\underbrace{(-r-2 \sin a)}_w \ = \ 0\tag{3}$$

Now, a turning point. Instead of using (3) in order to find the coordinates of $C$ and $D$, then find the 2 tangents to the parabola, and at last obtain the coordinates of their point of intersection $M$, there is a rather spectacular shortcut.

This shortcut is based on the fact that $M$-line $CD$ are a so-called "pole-polar" pairing (in the present case, being given a point external to a conic curve, one considers the two tangent lines to this conic curve ; the line joining the two tangency points is called the polar of this pole). We have a nice property which says that if

$$P=\begin{pmatrix}1&0&0\\0&0&-1\\0&-1&1\end{pmatrix}\tag{4}$$ is "the" matrix associated with conic $C_1$ (see explanations in Remark 1 below), we have the following relationship between the coordinates of $(x_M,y_M)$ of pole/point $M$ and coordinates $(u,v,w)$ of its polar line;

$$\underbrace{\begin{pmatrix}u\\v\\w\end{pmatrix}}_{\text{polar line}}\approx\underbrace{\begin{pmatrix}1&0&0\\0&0&-1\\0&-1&1\end{pmatrix}}_P \ \underbrace{\begin{pmatrix}x_M\\y_M\\t\end{pmatrix}}_{\text{pole = point}}\tag{5}$$

for a certain $t$ where the $\approx$ symbol means "up to a multiplicative factor".

Said otherwise, relationship (5) is a remarkable connection between points and lines through a linear transformation "naturally associated with" the given conic.

It is not difficult to solve the system of equations generated by (5), giving :

$$M=\begin{pmatrix}x_M\\y_M\\t\end{pmatrix}\approx \begin{pmatrix}\cos a \\r + \sin a\\ \sin a \end{pmatrix}\tag{6}$$

From (6), in order to show that $O,M,F$ are aligned, it is enough to show that the determinant of their homogeneous coordinates is null:

$$\begin{vmatrix}0&\cos a&r \cos a\\0&r + \sin a&2+r \sin a\\1&\sin a&1\end{vmatrix}=0$$

Remark 1: the symmetric matrix $P$ associated with conic curve is due to this factorization :

$$ax^2+2bxy+cy^2+2dx+2ey+f=0= \begin{pmatrix}x&y&1\end{pmatrix}\underbrace{\begin{pmatrix}a&b&d\\b&c&e\\d&e&f\end{pmatrix}}_P\begin{pmatrix}x\\y\\1\end{pmatrix}$$

Remark 2: In fact, the 2 conic curves are part of a so-called bitangent pencil of conics depending on a single parameter $\lambda$ with common equation :

$$\underbrace{(x-y)(x+y)}_U+\lambda \underbrace{(y-1)^2}_V=0 \ \ (conic \ curve \ C_{\lambda})\tag{2}$$

(check that for $\lambda=1$, one gets parabola $C_1$ and for $\lambda=2$, one gets circle $C_2$).

Remark in the remark : $U=0$ represents the pair of tangents (called a degenerated conic curve), $V^2=0$ represents another degenerated conic made of the equation of line $AB$ at power 2 because points $A, B$ are double points).

Remark 3 : Here is a certain number of references about the so-called bi-tangent pencils: See here (see Fig. 76 and 77 p. 335) and here. See as well an application here.

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    $\begingroup$ Sorry, it actually took some time to go through all the equations and the paper. It is clear what you were doing and the solution is interesting and very motivated. It did give us some insights to better understanding polars and poles. We also looked into the theory a bit more and now everything is completely clear regarding projective transformations too(reading through the polars and poles we found projective transformations as a very close topic which seems as an easy way of generalising the solution quickly for arbitrary conics), they are such a powerful tool $\endgroup$ Oct 6, 2021 at 16:54
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    $\begingroup$ [+1] @Jean Marie: Thanks for the references. It's always good to know both the analytic and synthetic sides. I didn't realize that 'bitangent' was another term for 'double contact'. Here's another Pamfilos paper on the topic: forumgeom.fau.edu/FG2009volume9/FG200922.pdf. Once upon a time I thought that double contact conics were an uninteresting edge case, but there are lots of pretty theorems about them, eg mathworld.wolfram.com/DoubleContactTheorem.html. (See also mathoverflow.net/questions/317514/a-problem-of-four-conics) $\endgroup$
    – brainjam
    Oct 7, 2021 at 17:43
  • $\begingroup$ @brainjam Thank you very much. $\endgroup$
    – Jean Marie
    Oct 7, 2021 at 22:17

In the projective plane, given a conic $C$, we have a mapping from points to lines, and from lines to points (polarity). But now let's say we have two conics $C_1$, and $C_2$. Take $P$ a points, $p$ its polar wr to $C_1$, and $Q$ the pole of $p$ wr to $C_2$. We got ourselves a map from points to points, and it is a projective map.

Now, assume moreover that $C_1$, $C_2$ have a double contact. They could be ( after a (complex) projective transformation) considered to be two concentric circles. Indeed, any circle passes through the circular points at infinity. If we have two circles, then they have a common chord formed by them. If moreover the circles are concentric, then that chord is of double tangency ( check in projective coordinates). The common tangents at these infinite cyclic points intersect in the common center. So we can reduce to the case of concentric circles (of equation $x^2 + y^2 = r_k^2$, $k=1,2$. In this case the map described above $P\to Q$ is a composition of inversions in the circles with center $O$ . This can be given a projective interpretation. In particular, it follows that $P$, $O$, $Q$ are collinear.

$\bf{Added:}$ concentric circles

The maps is $P\mapsto Q$ obtained as follows: the polar of $P$ with respect to the circle $c_1$ is the yellow line $l$, and the pole of $l$ with respect to $c_2$ is the point $Q$. In the figure $c_1$ and $c_2$ are concentric circles. If $d$ is the distance from the common center $O$ to the line $l$, then $d\cdot OP= r_1^2$, and $d\cdot OQ= r_2^2$, so $\frac{OQ}{OP} =\frac{r_2^2}{r_1^2}$, so the map $P\mapsto Q$ is a homothety with center $O$. In particular we conclude that $O$, $P$, $Q$ are collinear. Also note that $P\mapsto Q$ is the composition of the maps $P\mapsto R$ and $R\mapsto Q$. These latter of maps can be given a projective interpretation as a reflexion across the conics $c_1$, $c_2$ with center $0$ ( in this case when $c_i$ are circles, they are inverseions)

Note: The circular points at infinity were introduced by Poncelet. Also he gave a projective interpretation for the foci of a conic, involving these circular points. It turns out that in general a conic has $4$ foci, although a real conic only has two real foci ( and two complex conjugate ones). This ican simplifie some reasonings about conics

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    $\begingroup$ I hadn't thought to consider the circular points at infinity : very astute. Only a single remark : I have some difficulty to see the connection between map $P \rightarrow Q$ and "composition of inversions". $\endgroup$
    – Jean Marie
    Feb 11 at 8:34
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    $\begingroup$ @Jean Marie: Thank you! Added a picture and some details. The circular points were a revelation to me not too long ago. $\endgroup$
    – orangeskid
    Feb 12 at 7:53
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    $\begingroup$ Very clear now ! $\endgroup$
    – Jean Marie
    Feb 12 at 11:28

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