How to show that there exist no four circles as in the picture? These are distinct (their radii might differ or not) semicircles on the $x-axis$ as depicted below. I want to show that there exist no such four Euclidean semicircles such that the intersections occurs at a right angle as in the figure. How can show that algebraically?

 A: Here are three proofs, from the simplest to the more elaborate: 1) using hyperbolic geometry, 2) using analytical geometry, 3) using the Apollonius configuration.

*

*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 is  impossible (this sum of angles being twice the sum of angles of a triangle, itself less than $\pi$ as is well known for hyperbolic geometry).


*Algebraic proof:
Let us give indices $k=1...4$ to the circles from left to right in the order of their leftmost point, with resp. equations:
$$x^2+y^2-2a_kx-2b_ky+c_k=0$$
Let us recall that 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 centers of all these circles belong to the $x$ axis.
The four orthogonality conditions give rise to the following system:
$$\begin{cases}2a_1a_2&=&c_1+c_2&(A)\\2a_1a_3&=&c_1+c_3&(B)\\2a_3a_4&=&c_3+c_4&(C)\\2a_2a_4&=&c_2+c_4&(D)\end{cases}$$
Combining these equations under the form (A)-(B)+(C)-(D), one gets:
$$(a_1-a_4)(a_3-a_2)=0$$
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 would be identical, contradicting the fact that all the circles are assumed distinct.


*Using the Apollonius configuration:
[Following the suggestion of Alan Abraham] Let us consider the pencil of circles $\lambda C_1+(1-\lambda) C_4$ of circles generated by the two leftmost and rightmost circles ; the generic circle of this pencil has equation:
$$x^2+y^2-2(\lambda a_1+(1-\lambda)a_4)x+(\lambda c_1+(1-\lambda)c_4)=0$$
(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 $C_1 \cap C_4=\emptyset$, just exchange the roles of blue and red circles.
A: 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.
