What relation is between these two pictures featuring two families of orthogonal circles?

Question

What relation is between these two pictures featuring two families of orthogonal circles?

• Picture-I : Found this while searching about conic sections.

• Picture-II : Found this in this video.

Why the two figures are similar?

Answer 1

One of the references is from "Excursions in Geometry" by Charles Stanley Ogilvy (Dover, 1990) where it is explained using circles of Apollonius.

This figure features two families of circles

• Those passing through two fixed points, say $(-1,0)$ and $(1,0)$,

• Those orthogonal to each circle of the first family.

There are different ways to consider, and work, with this fascinating figure. Here are two of these treatments :

1. (Have a look to the right part of figure 1).

The blue circles have common equation

$$f(x,y):=x^2+y^2-2ax+1=0\tag{a}$$

where $(a,0)$ is the ordinate of their center (see Remark 3 below for an indication of proof).

The red circles have the characteristic property to pass through fixed points $A(1,0)$ and $B(-1,0)$. It is easy to show (exercice!) that their common equation is:

$$g(x,y):=x^2+y^2-2by-1=0\tag{b}$$

where $(0,b)$ is the ordinate of their center.

Let us show that any red circle is orthogonal to any blue circle. Let us add relationships (a) and (b), and multiply the result by 2:

$$4(x^2+y^2-ax-by)=0\tag{c}$$

(c) can be written:

$$2(x-a).2x+2y.2(y-b)=0$$

$$\Leftrightarrow \ \ \ \binom{2(x-a)}{2y}.\binom{2x}{2(y-a)}=0$$ $$\Leftrightarrow \ \ \binom{\partial f/\partial x}{\partial f/\partial y} \ \perp \ \binom{\partial g/\partial x}{\partial g/\partial y}\tag{d}$$

The gradients of $f$ and $g$ are orthogonal at intersection point $(x,y)$, therefore giving the orthogonality of the corresponding circles.

2. These two families of circles can be obtained in a different, more global, way by using complex functions geometry, under the point of view of "conformal transformations" (conformal = angle preserving). This is theoretically important, but as well practically useful because it provides an easy way to plot at once the two families of circles by considering a deformation (a "transformation") of a square grid.

Fig. 1 : Left: the initial plane of variable $w$ ; right: the "image plane" of variable $z$ with conformal transformation (1) "mapping" ("bending"...) the first set of orthogonal lines onto the second set of orthogonal circles.

The transformation is encapsulated into the following function

$$w \in \mathbb{C} \mapsto z=\dfrac{e^{w}+1}{e^{w}-1} \in \mathbb{C} \tag{1}$$ which is differentiable, therefore transforms a certain set of curves (here straight lines) into another set of curves (here circles) while preserving their ($\pi/2$) angles.

Remarks :

1. There are other ways to obtain a similar figure, this time with the parent transformation:

$$w \in \mathbb{C} \mapsto z=\tan w \in \mathbb{C} \tag{2}$$

evoking some Earth representation (with meridians orthogonal to parallel lines) or the field lines generated by a magnet...

Fig. 2 : The effect of transformation (2). This figure is plainly Fig. 1 with a $\pi/2$ rotation. This isn't in fact surprizing because replacing $w$ by $iw$ in (1) gives the formula for a (co)tangent (up to a replacement of $w$ by $w/2$ and multiplication by $i$)

2. See example 6 of this document which explains how (a) can be derived from (b) by solving a certain differential equation.

3. Transformation (1) can be written :

$$x+iy \ \mapsto \ z=\dfrac{re^{iy}+1}{re^{iy}-1} \ \ \text{with} \ \ r:=e^{x}$$

where we recognize in $re^{iy}$ the polar representation of a complex number. In this way, we see that the underlying basic transformation is :

$$w \ \mapsto \ z=\dfrac{w+1}{w-1}$$

4. The family of blue circles can be understood in the following way (Apollonius' approach). Any of them can be defined as the set of points $M$ such that the ratio $\dfrac{MA}{MB}$ is constant for a given circle. This is connected to the previous remark, because $\dfrac{MA}{MB}=\left|\dfrac{w+1}{w-1}\right|$ (see the reference I gave at the beginning).

5. Constant coefficients $1$ or $-1$ in equations (a) and (b) are called the power of the origin with respect to each family of circles. It is an efficient tool when working on circles.

6. See also Fig. 2 there.

Here is the Matlab program that has generated Fig. 1 :

   f=@(x,y)((exp(x+i*y)+1)./(exp(x+i*y)-1));%expression (1)
   x=-2:0.01:2;y=-pi:pi/100:pi;%range of x and y values
   subplot(1,2,1);hold on;% left fig. 
   for k=-3:0.21:3.;plot(x+i*k,'r');end;%red lines
   for k=-2:0.51:2.;plot(k+i*y,'b');end;%blue lines
   subplot(1,2,2);hold on;axis equal;% right fig. 
   axis([-2,2,-2,2]);
   for k=-3:0.2:3;plot(f(x,k),'r');end;%red circles
   for k=-2:0.5:2.1;plot(f(k,y),'b');end;%blue circles
   plot([-2,2],[0,0],'k');%hor. axis
   plot([0,0],[-2,2],'k');%vert. axis