TITLE: How to show that there exist no four circles as in the picture?
QUESTION [6 upvotes]: 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?

REPLY [4 votes]: 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.