The equation of the circumcircle of an equilateral triangle $\text{ is }{\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$ and one vertex of the triangle is (1, 1). The equation of the incircle of the triangle is

In an equilateral triangle, circumcenter and incenter are coincident. 

Therefore,$\begin{array}{rcl}\text{ Incenter }& \equiv & (-\mathrm{g},-\mathrm{f})\end{array}$

$\begin{array}{rcl} & & Since \text{ Point }(1,1)\text{ lies on the circle. }\end{array}$

$\begin{array}{rcl}{1}^2+1^2+2\mathrm{g}+2\mathrm{f}+\mathrm{c}& =& 0\\ \text{ or }\mathrm{c}& =& -2(\mathrm{g}+\mathrm{f}+1)\end{array}$

$\begin{array}{rcl} & & \mathrm{Also}, \mathrm{in} \mathrm{an} \mathrm{equilateral} \mathrm{triangle},\end{array}$

$\begin{array}{rcl}\text{ Circumradius }& =& 2\times \text{ Inradius }\\ \therefore \text{ Inradius }& =& \frac{1}{2}\times \sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}\end{array}$

$\begin{array}{rcl}\mathrm{Therefore}, \mathrm{the} \mathrm{eqlration} \mathrm{of} \mathrm{the} \mathrm{incircle} \mathrm{is} (\mathrm{x}+\mathrm{g}{)}^2+(\mathrm{y}+\mathrm{f}{)}^2& =& \frac{1}{4}\left({\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}\right)\\ & =& \frac{1}{4}\left\{{\mathrm{g}}^2+{\mathrm{f}}^2+2(\mathrm{g}+\mathrm{f}+1)\right\}\\ 4\left(x^2+y^2\right)+8gx+8fy& =& \left(1-g\right)\left(1+3g\right)+\left(1-f\right)\left(1+3f\right)\end{array}$