Given a circle C on the plane, what are the circles orthogonal to the x-axis and tangent to C?

Question

Let $C$ be a generalised circle in $\mathbb{CP}^1$ represented by a $2 \times 2$ indefinite Hermitian matrix also called $C$. (See here for how an indefinite Hermitian matrix determines a generalised circle). Given two points on the $x$-axis $x_1$ and $x_2$, denote by $C(x_1,x_2)$ the circle passing through $x_1$ and $x_2$ which is orthogonal to the $x$-axis, i.e. $C(x_1,x_2) = \{x + iy \mid \left(x-\frac{x_1+x_2}{2}\right)+y^2 = \left(\frac{x_1 - x_2}2\right)^2\}$. Is there a simple equation for the set of pairs $(x_1,x_2)\in \mathbb{RP}^1\times\mathbb{RP}^1$, such that that for each such pair, $C(x_1,x_2)$ is tangent to $C$?

My work: If I assume that I know the centre point of $C$, then I can find an implicit equation for set of all $C(x_1,x_2)$. If the centre point of $C$ is $(x,y)$ and the radius is $r$ then $r + |\frac{x_1 + x}{2}| = |\frac{x_1 + x}{2} - (x + iy)|$. But in general, $C$ is a generalised circle, so I can't assume that.

Answer 1

I can't think of a good solution which would use the complex plane to some advantage, since that setup is best suited to describing Möbius geometry and its focus on circles orthogonal to one another. If you need tangentiality, I would instead switch to Lie sphere geometry following the approach I detailed in another answer.

Switching from Möbius geometry to Lie geometry

One way to look at it, in Möbius geometry you are dealing with homogeneous coordinate vectors of the form $(x,y,x^2+y^2-r^2,1)$ to describe a circle with center $(x,y)$ and radius $r$, or $(a,b,-2c,0)$ to describe the line $ax+by+c=0$. You should be able to read these values (or any multiple thereof) from your Hermitian matrix representation, without needing a case distinction between circle and line. I'll add a section on that a bit further down.

Compared to that, Lie geometry has an additional entry for the unsquared radius $r$. Which you need to compute from the above in a way that's compatible with homogeneous coordinates, i.e. takes scalar multiples correctly into account. If you have a vector $v$ describing a generalised circle in the order mentioned above, then the corresponding field representing the radius would be $\sqrt{v_1^2+v_2^2-v_3v_4}$. Try it out for the vectors above: for the circle you get $r$, for a multiple of that circle vector you get the same multiple of $r$, and for the line you get $\sqrt{a^2+b^2}$ just as my other post claims.

Characterizing tangentiality

So now you can turn the circle $C$ into a five-element vector, and you can also turn your $C(x_1,x_2)$ into such a vector. And following the machinery from that other post you now can use $C\cdot L\cdot C(x_1,x_2)=0$ as condition for tangentiality.

Note that Lie geometry, contrary to Möbius geometry, deals with oriented circles. So you get oriented contact, and if you want to capture all tangential solutions, you want to also re-do the equations with the sign for the radius entry flipped in one of the two generalised circle vectors.

One way of achieving this is picking $\frac{x_1-x_2}2$ as the radius of $C(x_1,x_2)$ so that one order of points corresponds to positive sign and the opposite order to negative sign. This turns the problem of that radius sign into an advantage.

Switching from Hermitian matrix to Möbius vector

So let me focus on how to translate from your Hermitian matrix to a Lie geometry vector. According to your reference a point $z=x+iy$ is on circle $C$ if

\begin{align*} (z,1)\cdot C\cdot\begin{pmatrix}\bar z\\1\end{pmatrix}&=0 \\ (x+iy,1)\cdot\begin{pmatrix}a&b-ic\\b+ic&d\end{pmatrix}\cdot\begin{pmatrix}x-iy\\1\end{pmatrix}&=0 \\ a(x+iy)(x-iy)+(b-ic)(x+iy)+(b+ic)(x-iy)+d&=0 \\ a\left(x^2+y^2\right)+2bx+2cy+d&=0 \\ x^2+2\frac bax+\frac{b^2}{a^2}+y^2+2\frac cay+\frac{c^2}{a^2} &=\frac{b^2}{a^2}+\frac{c^2}{a^2}-\frac da \\ \left(x+\frac ba\right)^2 + \left(y+\frac ca\right)^2 &=\frac{b^2}{a^2}+\frac{c^2}{a^2}-\frac da \end{align*}

Now let me shift nomenclature from $(x,y)$ being a point on the circle to $(x,y)$ denoting the center of the circle as I did above. Reading the circle parameters from the equation above, we get center $x=-\frac ba$ and $y=-\frac ca$ and also $r^2=\frac{b^2}{a^2}+\frac{c^2}{a^2}-\frac da$ so $x^2+y^2-r^2=\frac da$. Thus

$$ \left(x,y,x^2+y^2-r^2,1\right) =\left(-\frac ba,-\frac ca,\frac da,1\right) \sim\left(-b,-c,d,a\right) $$

is the vector you could use, with the right formulation suitable for the case of $a=0$ which is not actually a circle.

Putting it all together

So if your circle is

$$C=\begin{pmatrix}a&b-ic\\b+ic&d\end{pmatrix}$$

then as discussed above the vector in my specific convention for Lie geometry would be

$$v=(-b,-c,\sqrt{b^2+c^2-ad},d,a)$$

where you might recognize the determinant occurring there in the square root.

For $C(x_1,x_2)$ you have a center at $x=\frac{x_1+x_2}2$ and $y=0$ with a signed radius of $r=\frac{x_1-x_2}2$ so you get $x^2+y^2-r^2=x_1x_2$. We can multiply all entries of the Lie vector by $2$ to avoid fractions and then we get

$$w=(x_1+x_2, 0, x_1-x_2, 2x_1x_2, 2)\;.$$

These two are tangential when

$$0=v\cdot L\cdot w^T=2b(x_1+x_2)+2\sqrt{b^2+c^2-ad}(x_1-x_2)+2ax_1x_2+2d$$

which you can simplify slightly by canceling a factor of two.