Hello, furnis1!

First, make a sketch . . .

[I'll use Calculus to maximize OC.]

A semi-circle is dawn outwardly on a chord $\displaystyle AB$ of a circle with center $\displaystyle O$ and raidus 1.

The point $\displaystyle C$ on this semi-circle is the furthest from center $\displaystyle O.$

Of course, the length of $\displaystyle OC$ is dependant on the choice of chord $\displaystyle AB.$

Determine $\displaystyle AB$ so that $\displaystyle OC$ is maximum. Code:

* * *
* * A
* o
* / :* o
1/ : o
* / θ D: *
* O * - - - + * - - o C
* \ : *
1\ : o
* \ :* o
* o
* * B
* * *

The circle with center $\displaystyle O$ has radii $\displaystyle OA = OB = 1.$

A semicircle is constructed (outward) with center $\displaystyle D$ and diameter $\displaystyle AB.$

. . It has radii: $\displaystyle DA = DC = DB.$

Let $\displaystyle \angle AOC = \theta,\;\;OC = d$

In right triangle $\displaystyle ADO\!:\;\;OD = \cos\theta\,\text{ and }\,DA = \sin\theta = DC$

Then: .$\displaystyle d \:=\:OD + DC \:=\:\cos\theta + \sin\theta$

To maximize $\displaystyle d$, set $\displaystyle d' = 0$

We have: .$\displaystyle d'\:=\:\sin\theta + \cos\theta \:=\:0\quad\Rightarrow\quad \sin\theta \:=\:\cos\theta \quad\Rightarrow\quad \tan\theta\:=:1$

. . Hence: .$\displaystyle \theta \:=\:45^o$

$\displaystyle OC$ is maximum when $\displaystyle \theta = 45^o$

Then $\displaystyle \angle AOB = 90^o$ and $\displaystyle \Delta AOB$ is an isosceles right triangle.

Since its legs are length 1, its hypotenuse is: .$\displaystyle AB \:=\:\sqrt{2}$