Circle covers square 2

Let the center of the circle be $O$ , its radius be $r$ , and the angle between the end of the arc at the top edge and the left edge of the unit square be $\theta$ as shown in the figure. Then the area of the orange region is given by:

$\begin{aligned} \overbrace{\frac {(\pi - \theta)r^2}2}^{\text{Area of sector}} + \overbrace{\frac {r^2\cos \theta \sin \theta}2}^{\text{Area of triangle}} & = \frac 12 & \small \color{#3D99F6} \text{Note that }\cos \theta = \frac {1-r}r \implies \theta = \cos^{-1} \left(\frac {1-r}r \right) \\ \left(\pi - \cos^{-1}\left(\frac {1-r}r\right)\right)r^2 + (1-r)\sqrt{2r-1} & = 1 & \small \color{#3D99F6} \implies \sin \theta = \frac {\sqrt{r^2-(1-r)^2}}r =\frac {\sqrt{2r-1}}r \end{aligned}$

Solving it numerically, we have $r \approx \boxed{0.596}$ .

Let $O$ be the center of the circle, $A$ be the corner of the square in the circle, $C$ be the corner of the square that is on the circle, $B$ be the other intersection of the square and the circle, $r$ be the radius of the circle, and $\theta$ be $\angle AOB$ .

From $\triangle AOB$ we find that $AB = \sqrt{2r - 1}$ and $\theta = \cos^{-1} (\frac{1 - r}{r})$ , so $\triangle AOB$ has an area of $A_{\text{triangle}} = \frac{1}{2}(1 - r)\sqrt{2r - 1}$ .

This means $\angle BOC = \pi - \theta = \pi - \cos^{-1} (\frac{1 - r}{r})$ , so sector $BOC$ has an area of $A_{\text{sector}} = \frac{\pi - \cos^{-1} (\frac{1 - r}{r})}{2\pi}\pi r^2$ .

The part of the circle covering the square is $\triangle AOB$ and sector $BOC$ , which must have a combined area of $A_{\text{triangle}} + A_{\text{sector}} = \frac{1}{2}$ .

Therefore, $\frac{1}{2}(1 - r)\sqrt{2r - 1} + \frac{\pi - \cos^{-1} (\frac{1 - r}{r})}{2\pi}\pi r^2 = \frac{1}{2}$ , which solves numerically to $r \approx \boxed{0.5962}$ .

First, place the unit square on the coordinate plane, with vertices at $(0,0), (1,0), (1,1), (0,1)$ . Then, place the circle with the both point at $(0,0)$ . The graph for this circle is $x^{2}+(y- \sqrt{a})^{2}=a$ , where $\sqrt{a}$ is the radius of the circle. If the circle has a radius of $\frac{1}{2}$ , the top of the circle will be at one of the vertices of the square, and the point furthest right on the circle hits the midpoint of the square. Because of the curve of the circle, this is obviously less than half of the square. So therefore $\sqrt{a} \gt \frac{1}{2}$ , and $a \gt \frac{1}{4}$ . Then, take the integral from $1$ to $0$ , with respect to y, of the circle. Rearranging the equation of the circle we get $x=\sqrt{a-(y-\sqrt{a})^2}$ . In integral format, we get $\int_{0}^{1} \sqrt{(a-(y-\sqrt{a})^2} dy$ , and we want this to equal $\frac{1}{2}$ , half of the area of the unit square. I tried various numbers on desmos, until I found that $a=0.355485596577$ , and so the radius is $\sqrt{a}=0.596226128727$ .