Find the radius of the circle cutting sides of equilateral triangle Consider the figure below. The triangle is equilateral. Find the radius of the circle.

My try:
I have dropped perpendiculars from the center of the circle to respective chords naming them $x,y,z$ and named the other line segments as $m,n,p,q,s,t$ as shown below.

By Viviani's theorem we have
$$x+y+z=\frac{\sqrt{3}a}{2}$$ Where $a$ is the side length of the triangle.
By Chords theorem we have:
$$\begin{aligned}
m(m+5) &=t(t+3) \\
q(q+4) &=s(s+3) \\
p(p+4) &=n(n+5) \\
m+n+5 &=p+q+4=s+t+3=a
\end{aligned}$$
Also by Pythagoras theorem if $R$ is the radius of the circle we have:
$$R^{2}-x^{2}=\frac{9}{4}, \quad R^{2}-y^{2}=4 ,\quad R^{2}-z^{2}=\frac{25}{4}$$
I guess this is a very tedious task? Any better ways?

If you set up a Cartesian coordinate system with its origin at the center of the circle and $BC$ as a horizontal line beneath it, then the coordinates of the chord endpoints work out to:
$$G = \left( - \frac{5}{2}, - \sqrt{R^2 - \frac{25}{4}} \right)$$
$$T = \left( \frac{5}{2}, - \sqrt{R^2 - \frac{25}{4}} \right)$$
$$N = \left(1 + \frac{\sqrt{3}}{2}\sqrt{R^2 - 4}, \frac{1}{2}\sqrt{R^2 - 4}  - \sqrt{3} \right)$$
$$M = \left(-1 + \frac{\sqrt{3}}{2}\sqrt{R^2 - 4}, \frac{1}{2}\sqrt{R^2 - 4} +  \sqrt{3}\right)$$
$$E = \left(\frac{3}{4} - \frac{\sqrt{3}}{2} \sqrt{R^2 - \frac{9}{4}}, \frac{1}{2}\sqrt{R^2 - \frac{9}{4}} + \frac{3\sqrt{3}}{4} \right)$$
$$F = \left(-\frac{3}{4} - \frac{\sqrt{3}}{2} \sqrt{R^2 - \frac{9}{4}}, \frac{1}{2}\sqrt{R^2 - \frac{9}{4}} - \frac{3\sqrt{3}}{4}\right)$$
From these, we can find equations for the lines that form the three sides of the triangle.
$$\overline{AB} = \overline{MN} : y = -\sqrt{3}x + 2\sqrt{R^2 - 4}$$
$$\overline{AC} = \overline{EF} : y = \sqrt{3}x + 2\sqrt{R^2 - \frac{9}{4}}$$
$$\overline{BC} = \overline{GT} : y = -\sqrt{R^2 - \frac{25}{4}}$$
From the intersections between those three lines, we find the triangle's vertices.
$$A = \overline{AB} \cap \overline{AC} = \left( \frac{1}{\sqrt{3}}\sqrt{R^2 - 4} - \frac{1}{\sqrt{3}}\sqrt{R^2 - \frac{9}{4}}, \sqrt{R^2 - 4} + \sqrt{R^2 - \frac{9}{4}} \right)$$
$$B = \overline{AB} \cap \overline{BC} = \left( \frac{1}{\sqrt{3}}\sqrt{R^2 - \frac{25}{4}} + \frac{2}{\sqrt{3}}\sqrt{R^2 - 4}, -\sqrt{R^2 - \frac{25}{4}} \right)$$
$$C = \overline{AC} \cap \overline{BC} = \left(-\frac{1}{\sqrt{3}}\sqrt{R^2 - \frac{25}{4}} - \frac{2}{\sqrt{3}}\sqrt{R^2 - \frac{9}{4}}, -\sqrt{R^2 - \frac{25}{4}} \right)$$
And by applying the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ to any two of these points, the common side length $a$ of the equilateral triangle is:
$$a = \frac{2}{\sqrt{3}}\left( \sqrt{R^2 - \frac{9}{4}} + \sqrt{R^2 - 4} + \sqrt{R^2 - \frac{25}{4}} \right)$$
But this is just what OP already found by Vivani's theorem, and adds no new information.
As indicated by @math for entertainment's answer, there isn't a unique solution by $R$.  We need one more piece of information to find it.
Assuming that information is the side length $a$, then we'll need to solve the above radical equation for $R$.
