Find the radius of a circle that encloses $5$ squares of side length $c$

Question

Recently, a brand new user had asked a question (somewhat awkwardly and incompletely). In his case the square's side length was fixed $c=1$.

But if you look at the question in generalized terms, it becomes actually interesting. Due to the downvotes, the user unfortunatelly has withdrawn his question. I find it a pity and therefore give it a second attempt by reformulating and generalizing the original question.

As shown in the picture below, a circle encloses $5$ squares of the length $c$ and I added some basic ideas:

Let us search the formula for the radius of the enclosing circle. At the first glance, the center of the circle seems to divide the square's side in a ratio of $2:3$. Is this assumption correct?

Answer 1

The picture shows that $$c^2+(c+d)^2=r^2=(\tfrac c2)^2+(2c-d)^2.$$ This implies that $d=\tfrac{3}{8}c$, so the center of the circle divides the square's side in a ratio of $3:5$, and $$r=\frac{\sqrt{185}}{8}c.$$

Answer 2

We just need to find a radius of the circumcircle for the triangle with sides-lengths $2c$, $\frac{3}{2}c\sqrt5$ and $\frac{c\sqrt{37}}{2}.$

Can you end it now?

I got the following.

We see that the area of the triangle it's $\frac{2c\cdot3c}{2}=3c^2$ and we obtain: $$R=\frac{2c\cdot\frac{3}{2}c\sqrt5\cdot\frac{c\sqrt{37}}{2}}{4\cdot3c^2}=\frac{c\sqrt{185}}{8}.$$

Answer 3

Another way.

We have the following system:

$$a+b+3c=2R,$$ $$c^2=b(2R-b)$$ and $$\frac{c^2}{4}=b(2R-b).$$ From the second and third we obtain: $$\frac{3c^2}{4}=(a-b)(2R-a-b)$$ or $$\frac{3c^2}{4}=(a-b)3c,$$ which gives $$b=a-\frac{c}{4},$$ which with $a+b+3c=2R$ gives $$a=R-\frac{11c}{8}$$ and we obtain: $$c^2=\left(R-\frac{11c}{8}\right)\left(R+\frac{11c}{8}\right)$$ or $$R=\frac{c\sqrt{185}}{8}.$$