How to show that there exist no four circles as in the picture?

Preliminary note: I have edited my answer because I have discovered a very direct proof (section 1)

Here are three proofs: 1) using hyperbolic geometry, 2) using analytical geometry, 3) using the Applonius configuration.

1. With hyperbolic geometry:

Using the [Poincaré half-plane model] (http://pi.math.cornell.edu/~mec/Winter2009/Mihai/section6.html), the upper part of the figure displays an "ordinary" quadrilateral whose sum of angles is $2 \pi$ which isn't possible (this sum of angles being twice the sum of angles of a triangle, itself less than $\pi$ as is well known for hyperbolic geometry).

2. Algebraic proof:

Let us give index $k=1...4$ to the circles from left to right in the order of their leftmost point, with resp. equations:

$$x^2+y^2-2a_kx-2b_ky+c_k=0$$

In general, a necessary (and in fact sufficient) condition for circles with equations:

$$\begin{cases}x^2+y^2-2ax-2by+c&=&0\\x^2+y^2-2a'x-2b'y+c'&=&0\end{cases}\tag{0}$$ to be orthogonal is

$$2(aa'+bb')=c+c'$$

(Reference: https://mathworld.wolfram.com/OrthogonalCircles.html).

Please note that $(a,b)$ and $(a',b')$ are the coordinates of the centers of these circles. Here, the $b,b'$ components are zero because the circles have their centers on the $x$ axis.

Therefore, the four orthogonality conditions give rise to the following system:

$$\begin{cases}2a_1a_2&=&c_1+c_2&(A)\\2a_1a_3&=&c_1+c_3&(B)\\2a_3a_4&=&c_3+c_4&(C)\\2a_2a_4&=&c_2+c_4&(D)\end{cases}$$

$$(A)+(C) \ \text{give} \ \ 2(a_1a_2+a_3a_4)=c_1+c_2+c_3+c_4\tag{1}$$

$$(B)+(D) \ \text{give} \ \ 2(a_1a_3+a_2a_4)=c_1+c_2+c_3+c_4\tag{2}$$

In this way, we can eliminate variables $c_k$ by equating (1) and (2), and obtain:

$$a_1a_3+a_2a_4-a_1a_2-a_3a_4=0 \ \iff \ (a_1-a_4)(a_3-a_2)=0$$

As $a_1=a_4$ is an excluded case (according to the figure), we must have $a_2=a_3$.

Replacing now $a_3$ by $a_2$ into relationship (B), and comparing with (A), we get $c_2=c_3$ which would mean that the 2nd and 3rd circle, having $a_2=a_3$ and $c_2=c_3$, would have the same equation, therefore be identical, contradicting the fact that all the circles are assumed distinct.

3. Using the Apollonius configuration:

Following the suggestion of Alan Abraham, let us consider the leftmost and the rightmost circles $C_1$ and $C_4$. The pencil of circles generated by these circles is either (obtained by adding $\lambda$ times the equation of $C_1$ and $(1-\lambda)$ times the equation of $C_4$), giving :

$$x^2+y^2-2(\lambda a_1+(1-\lambda)a_4)x+(\lambda c_1+(1-\lambda)c_4)=0$$

(with all their centers on the $x-axis). In the case of the given figure, $C_1 \cap C_4 = {A,A'}$, therefore the pencil above is the set of red circles in the figure below:

Fig. 1: the two families of orthogonal circles.

The set of circles orthogonal to $C_1$ and $C_4$ is a certain set of circles (in blue on figure above) with their centers on line $AA'$ and the property that any point on $AA'$ is the center of a well determined blue circle ; as we are looking for a center on the $x$ axis, there is only one such circle. Let it be circle $C_2$. In this case, there is no place for a circle $C_3$ different from $C_2$ (with its center on the $x$ axis...

Remark: if $C_1 \cap C_4=\emptyset$ reverse the role of blue and red circles.

Let's simplify the problem by an inversion. Specifically, let's invert about a right-angle intersection between one of the two semicircles.

Those two semicircles (or rather the circles they represent) becomes a pair of perpendicular lines; without loss of generality, they're the $x$-axis and the $y$-axis.

The line in your diagram becomes a circle perpendicular to both axes, which means its center must be at the origin; without loss of generality, it's the unit circle.

For the remaining two semicircles (or rather the circles they represent):

• One must be perpendicular to the $x$-axis, which means its center has coordinate $(a,0)$ for some $a$. Without loss of generality $a>0$.
• The other must be perpendicular to the $y$-axis, which means its center has coordinate $(0,b)$ for some $b$. Without loss of generality $b>0$.
• Both must be perpendicular to the unit circle. If the circle centered at $(a,0)$ intersects the unit circle at $(x,y)$, then the three points $(0,0)$, $(a,0)$, and $(x,y)$ form a right triangle with hypotenuse $a$ whose other two sides are the radii. So the radius of this circle is $\sqrt{a^2-1}$. Similarly, the circle centered at $(0,b)$ has radius $\sqrt{b^2-1}$.

For a circle with radius $\sqrt{a^2-1}$ to be orthogonal to a circle with radius $\sqrt{b^2-1}$, the distance between their centers must be $$ d = \sqrt{(\sqrt{a^2-1})^2 + (\sqrt{b^2-1})^2} = \sqrt{a^2 + b^2 - 2}. $$ But we already know that the distance between their centers $(a,0)$ and $(0,b)$ is $\sqrt{a^2+b^2} \ne \sqrt{a^2+b^2-2}$, contradiction.