# Show that the locus of a point in this geometric construction is a conic

Start with a circle $$c$$ (black), conic $$d$$ (green) and a point $$A$$. $$K$$ is a point on the conic, and the tangent at $$K$$ intersects $$c$$ at $$F$$ and $$G$$. Line $$GH$$ is perpendicular to $$AG$$ and line $$FH$$ is perpendicular to $$AF$$.

The question: Show that the locus $$e$$ of $$H$$ (red) as $$K$$ moves on $$d$$ is a conic.

An observation: the points $$A,G,F,H$$ are concyclic. This suggests that the construction can be run in reverse. Start with conic $$e$$ instead of $$d$$. Let $$H$$ be a point on $$e$$ and let $$G,F$$ be the intersection of circle $$c$$ with the circle whose diameter is $$AH$$. Then the line $$GH$$ envelopes a conic as $$H$$ runs through $$e$$.

I've unsuccessfully tried a few things, such as

• showing that $$e$$ is the intersection of two projective pencils
• showing that the conic cross ratio is preserved by the mapping $$K\to H$$ (it appears to be preserved)
• simplifying by assuming that $$d$$ is a circle.

The problem/question given here is a simplification of an earlier math.SE question about a circumcenter locus. That question is answered with a (self described) brute-force vector proof. Although that proof should easily translate to address the present question, I was hoping for something simpler.

I'd prefer a synthetic proof, but new analytic proofs are welcome. Also welcome is any reference to a similar or identical problem and its proof (most likely in a text published in the period 1875-1925).

• Where does this question comes from ? Oct 29, 2021 at 10:37
• @Jean Marie, that's often my first comment! I've modified the question to give this info. Oct 29, 2021 at 14:40
• Thank you very much. Oct 29, 2021 at 19:12

Starting with an observation: If $$A$$ were the center of $$c$$, then $$H$$ would be the pole of line $$FG$$ with respect to that circle. In that case, the fact that conics map to conics could be easily seen in an analytic description from the fact that a polar-to-pole transformation is only a matrix multiplication.

Maybe that is something we can show in general? I'm afraid my way of doing so has a lot of analytic brute-force aspects to it, so it might not be exactly what you were hoping for, but perhaps it can still be of some use.

Let's start by assuming, without loss of generality, that $$c$$ is the unit circle, i.e. in homogeneous coordinates $$x^2+y^2-z^2=0$$. Let's denote your line $$FG$$ as $$g$$, and pick generic homogeneous coordinates for it and for the point $$A$$:

$$A=\begin{bmatrix}A_1\\A_2\\A_3\end{bmatrix}\qquad g=\begin{bmatrix}g_1\\g_2\\g_3\end{bmatrix}$$

The points where $$g$$ intersects $$c$$ I will call $$G_{1,2}$$; they correspond to your $$F,G$$.

$$G_{1,2}=\begin{bmatrix}-g_1g_3\\-g_2g_3\\g_1^2+g_2^2\end{bmatrix}\pm \lambda\begin{bmatrix}g_2\\-g_1\\0\end{bmatrix}\quad \text{with }\lambda:=\sqrt{g_1^2+g_2^2-g_3^2}$$

Now we need to construct $$H$$ from these $$G_i$$. Start by connecting $$G_i$$ to $$A$$ using a cross product. Take the point at infinity in the orthogonal direction by multiplying with

$$J:=\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}$$

Connect that point at infinity with $$G_i$$ again to get the perpendicular line in that point. And last intersect the two lines you obtained in this fashion. In formula:

\begin{align*} h_i&=J(A\times G_i)\times G_i \\ H&=h_1\times h_2 \end{align*}

If you feed this to a computer algebra system of your choice, the result may look massive at first glance. But there is a lot of common factors in there. Specifically I was able to identify a common factor of

$$\mu:=2(g_1^2+g_2^2)^3\sqrt{g_1^2+g_2^2-g_3^2}$$

The fact that this term must not be zero can be seen as a non-degeneracy constraint, requiring your line to be neither a tangent to $$c$$ nor the line at infinity (or some other line with complex coordinates passing through one of the ideal circle points). You can cancel that factor (and remove the non-degeneracy constraint) since we are dealing with homogeneous coordinates. The result can then be expressed as

$$H=\mu Mg\sim Mg\quad\text{with }M:=\begin{bmatrix} A_2^2-A_3^2 & -A_1A_2 & -A_1A_3 \\ -A_1A_2 & A_1^2-A_3^2 & -A_2A_3 \\ A_1A_3 & A_2A_3 & A_3^2 \end{bmatrix}$$

Remember that the above is w.l.o.g. assuming $$c$$ to be the unit circle. For other circles, the entries of $$M$$ would need to take your choice of circle into account.

So we can confirm that the map from $$g$$ to $$H$$ is a simple matrix multiplication. For the special case discussed initially, where $$A$$ is the center of the unit circle $$c$$ you get $$A_1=A_2=0$$ and $$M$$ becomes the matrix of the unit circle, which is exactly what you multiply with to convert polar to pole.

You can obtain $$g$$ from $$K$$ by multiplying with the primal matrix of $$d$$, so this shows that the map $$K\to H$$ is just a matrix multiplication as well and as such will preserve conic cross ratios, concluding one of your attempts for a solution.

And to be extra clear, here is how $$M$$ helps you to get one conic from the other. If you write $$D_d$$ as the dual matrix of conic $$d$$ (i.e. a line is tangent if the quadratic form becomes zero), and $$P_e$$ as the primal matrix of conic $$e$$ (i.e. a point is incident if the quadratic form becomes zero) then up to a scalar factor you get

$$H^TP_eH = (Mg)^TP_e(Mg) = g^T(M^TP_eM)g = g^TD_dg \\ D_d = M^TP_eM\qquad P_e = (M^T)^{-1}D_dM^{-1}$$

For non-degenerate conics you can use matrix inversion to convert between primal and dual representation of each conic.

Note that if $$A_1^2+A_2^2-A_3^2=0$$ or $$A_3=0$$ then $$\det(M)=0$$ so you would get a singular matrix which maps a non-degenerate conic $$d$$ to a degenerate conic $$e$$ in a way that can not be reversed. You might want to require $$A$$ to neither lie on $$c$$ nor on the line at infinity if you want to avoid this situation.

• Thanks! This is a very nice proof, very nicely explained, using well known quadratic form methods. Not at all brute force, and it suggests how a synthetic proof may go. Oct 31, 2021 at 17:40
• [+1] very very didactic proof. Nov 9, 2021 at 7:47
• A little comment: in the vein of an "wholy matricial solution", it should be underlined that cross-multiplication by a constant vector with entries $(a,b,c)$ is a linear operation associated with skew symmetric matrix $\begin{pmatrix}0&c&-b\\-c&0&a\\b&-a&0\end{pmatrix}$. Nov 9, 2021 at 8:01
• @JeanMarie: While true, I don't see how that matrix notation can help. To me it would be most useful if a matrix notation of everything else would lead directly to $M$, but the fact that $g$ occurs with degree $8$ and those annoying square roots in $H$ makes it really hard to keep matrix operations and input vectors separated along the way there. I thought maybe you saw the pattern of that cross product matrix in the off-diagonal part of $M$, but since $M_{12}=M_{21}$ share the same sign, that is not a good fit either.
– MvG
Nov 9, 2021 at 12:39