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

*

*(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.


*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 :

*

*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 $i$)


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


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


*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).


*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.


*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


