There is a geometric construction that I heard years ago and I still haven't figured out why it works despite several attempts. Playing with pen, paper and GeoGebra makes me confident that it does indeed work. Could someone explain it to me?

Task: Given a circle and a point outside it, construct the two tangents to the circle through the point, using only a straightedge.

Solution: Draw any three different lines through the given point $P$ that intersect the circle twice. Let $A_1,A_2,B_1,B_2,C_1,C_2$ be the six intersection points, with the same letter corresponding to the same line and the index 1 corresponding to the point closer to $P$. Let $D$ be the point where the lines $A_1B_2$ and $A_2B_1$ intersect, and similarly $E$ for the lines $B_1C_2$ and $B_2C_1$. Draw a line through $D$ and $E$. This line meets the circle at two points, $F$ and $G$. The tangents are $PF$ and $PG$.


6 Answers 6


It is a consequence of Pascal's theorem in Projective geometry valid for any conic, the circle is a special case. When separation tends to zero, the secant points $(OA_1, OA_2)$ where $(A_1, A_2) $ are extremities of chord on circle which multiply to tangent point $F$ as

$ OA_1 OA_2= OB_1 OB_2=...= OD_1 OD_2= PF^2$ as constant product, a property of the circle.

The locus,chord $ FG $, is a straight line. As a property of circle it is the harmonic mean of segments $ (OA_1, OA_2) $ as,for all chords e.g.,

$$ 2/PF = 1/OA_1 + 1/OA_2. $$


Pascal line is the chord of harmonic means of a pencil/circle intersection

Pascal Line Circle

Pascal Line valid on ellipse also, but I do not know if that is the harmonic locus. (posted Sept 2, 2016).

Pascal Line Ellipse

  • $\begingroup$ I'm not sure I understand your answer completely. I was previously unaware of Pascal's theorem and it seems useful. You say that the Pascal line corresponding to the choice of the three lines is the line $FG$. But the Pascal line could change if we change the three lines, so I don't see how the limit argument works exactly. It seems that the Pascal line is indeed independent of the choice of the three lines, but this requires an argument. And I don't see how the harmonic mean is relevant here. Can you elaborate? $\endgroup$ Aug 15, 2015 at 19:37
  • $\begingroup$ If $ A_1 $ is brought near to $ A_2$ by shifting up our straight edge it tends to position $F$.as a double point of tangency. Similarly with $C'$ s and $G$. Since harmonic mean leads to an invariant property chord line for all rays like $ PA_1A_2, ...,...$ the initial choice in relative positioning of $ A, B , C $ should not matter. $\endgroup$
    – Narasimham
    Aug 15, 2015 at 20:13
  • 1
    $\begingroup$ Added a hand made Paint sketch, poor accuracy, for sake of reference. The pole/polar discussion is also relevant here., $\endgroup$
    – Narasimham
    Aug 15, 2015 at 20:41

Suppose your construction is a perspective image (a central projection) of 'something' drawn on some plane $p$, with the point $P$ being on the horizon of the perspective, that is 'in infinity'. What is the 'original' drawing on the plane $p$ then?

Well, it is an ellipse, with three parallel lines $A,B,C$ crossing it.

Now, we can apply a shear transformation to the $p$ plane to transform the ellipse into a circle. A shear mapping preserves collinearity, parallelism and tangency, so the transformed image would be a circle with three parallel lines $A,B,C$ crossing it.

The respective chords of the circle are parallel, so they are bases of two trapezoids.
Due to the circle's symmetry, those trapezoids are isosceles and have a common axis of symmetry, and that axis goes through the circle's center point.
The diagonals of isosceles trapezoids cross on the axis of symmetry, so your $DE$ line is the same axis. It goes through the center of the circle and it is perpendicular to the $A,B,C$ lines
– so $FG$ is a diameter perpendicular to $A,B,C$, and its endpoints are tangency points for two lines paralel to $A,B,C$.

And those tangent lines meet $A,B,C$ in $P$ on the original construction due to properties of perspective mapping, so the construction is correct.

  • $\begingroup$ This sounds like a very interesting idea. Can you give more details about the transformation you apply to the original picture to move P to infinity? The idea is good, but I fail to reconstruct the exact transformation from your text. $\endgroup$ Sep 2, 2016 at 14:28

The short answer is that the points $E,F$ are on the polar $p$ of $P$ with respect to the circle, and the tangents from $P$ go through the points where $p$ intersects the circle. Moreover, this construction works for all conics, not just circles. And you can get away with a construction that uses fewer lines.

There are lots of 19th century projective geometry texts on archive.org that explain this, but I'll point you to Hatton's Projective Geometry, pg 132.

This illustration from that page shows the construction.

Hatton Polar

This construction gets away without computing $F$ because $Q$ is also on the polar. The proof for this construction is give in the two preceding pages of the reference, and actually starts from the construction you give (chords $AA',BB',CC'$).


Here is an excerpt from Kiselev's Geometry; Book I: Planimetry (English Adaptation) regarding tangents through a point:

Problem. Through a given point, to draw a tangent to a given circle.

Consider two cases:

(1) The given point ($C$) lies on the circle itself. Then draw the radius to this point, and at its endpoint $C$, erect the perpendicular $AB$ to this radius [...].

(2) The given point ($A$) lies outside the disk bounded by the given circle. Then, connecting $A$ with the center $O$, construct the circle with $AO$ as a diameter. Through the points $B$ and $B'$ at which this circle intersects the given one, draw the lines $AB$ and $AB'$. These lines are the required tangents, since the angles $OBA$ and $OB'A$ are right (as inscribed intercepting a diameter).

Note that the solution presented above does require a compass to draw a circle (i.e. the circle with diameter $AO$).

Here is an excerpt pertaining to the construction in (1) of the above excerpt:

Problem. To construct a right triangle given its hypotenuse $a$ and a leg $b$.

On a line $MN$, mark $AB = a$ and describe a semicircle with $AB$ as a diameter. (For this, bisect $AB$, and take the midpoint for the center of the semicircle and $\frac{1}{2}AB$ for the radius.) Then draw an arc of radius congruent to $b$ centered at point $A$ (or $B$). Connect the intersection point $C$ of the arc and the semicircle, with the endpoints of the diameter $AB$. The required triangle is $ABC$, since the angle $C$ is right, $a$ is the hypotenuse, and $b$ is a leg.

There is also another way of constructing right angles, which is more primitive that the excerpt above.

However, as mentioned already, these constructions require a compass in addition to a straightedge.

I'm not aware of a solution to your problem with solely a straightedge.

But, remember what a compass serves to do: find the locus of points equidistant from some point (i.e. draw a circle). Can we not do the same with a straightedge? Yes, and no. You can only find a finite amount with a straightedge, so it won't really be a locus. The compass merely draws every point with just a swipe of the wrist.

Let us "emulate" a compass with a straightedge:


Pick a point $A$ on your straightedge. Pick another point $B$ on your straightedge (such that $AB$ will be the radius of your desired circle). Put your pencil tip on this $B$ and start rotating the straightedge about $A$.

My answer is more of a hack, rather than a geometric proof. If your life depended on finding a tangent to a circle going through some point, and you only had a straightedge, I recommend doing what I said above. However, don't torture yourself by not having a compass handy if you don't need to. :)

  • $\begingroup$ In (1), how do you get a perpendicular just with a straightedge ? $\endgroup$
    – user65203
    Sep 1, 2016 at 7:19
  • $\begingroup$ In (2), how do you find the midpoint of the circle with $AO$ as a diameter without a compass, and how do you find the points $B$ and $B'$ without drawing the new circle? I don't see how I could actually draw the construction without a compass. $\endgroup$ Sep 1, 2016 at 7:50
  • $\begingroup$ @YvesDaoust : I'm not aware of such a possibility. However, see my updated answer and you can "emulate" a compass with a straightedge to construct (1) with just a straightedge. $\endgroup$
    – Fine Man
    Sep 1, 2016 at 19:04
  • $\begingroup$ @JoonasIlmavirta : I can't find a way not to use a compass. :). However, as I mentioned to YvesDaoust, see my updated answer to see how you can "emulate" a compass with a straightedge. It's more of a hack than a true answer, I suppose. :) $\endgroup$
    – Fine Man
    Sep 1, 2016 at 19:05
  • $\begingroup$ One would assume that a compass is indeed necessary, but the method I put in my question does not use a compass and it seems to work. I just don't see why, and that was my question. Oh, and that is a hack. :o) $\endgroup$ Sep 1, 2016 at 20:31


Harmonic mean $HM$ segments are of length $$ (OA_1, OA_2) = OH_A,\, (OB_1, OB_2) = OH_B,\,(OC_1, OC_2)= OH_C,..$$ measured from the pole $O$ always lie on the polar line represented by red circle locus. It contains points $ (H_A, H_B, H_C..) $ and end tangent points $ T_1 T_2 $ where the power of the circle is $ OT_1^2=OT_2^2$.

Pole & Polar

It is interesting that the neighboring segments can be used as well for this purpose as the points of intersection (black circles) continue to be on the polar, a fact interestingly relevant to draw chord line as tangent points $ T_1 T_2 $ connecting line.

  • $\begingroup$ It is continuation of an earlier answer (where I have mentioned about harmonic mean) can now be found another way. $\endgroup$
    – Narasimham
    May 12, 2020 at 18:45

The answer is because when you fix one line through $P$ that intersects the circle at $A_1$ and $A_2$ and vary a second line and hence its intersection points $B_1$ and $B_2$, the point $E$ defined by you varies on a line.

This should be fairly easy and boring to check with analytic geometry.

It bugs me too that I cannot see the reason in some ratios or some such though.

  • 1
    $\begingroup$ Have you actually checked that the calculation using analytic geometry is doable? Even if it is, it would be somewhat unsatisfactory, since there ought to be a proof "in a Euclidean spirit". $\endgroup$ Aug 15, 2015 at 17:11
  • $\begingroup$ HINT Harmonic mean of $ (OA_1, OA_2) $ ,.., .. always lies on (locus) $ FG $. $\endgroup$
    – Narasimham
    Aug 15, 2015 at 18:17

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