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

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$4 (x^{2} + y^{2}) + 8 gx + 8 fy = g^{2} + f^{2}$

$4 (x^{2} + y^{2}) = g^{2} + f^{2}$

$4 (x^{2} + y^{2}) + 8 gx + 8 fy = (1 - g) (1 + 3 g) + (1 - f) (1 + 3 f)$

none of these

answer is B.

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Detailed Solution

In an equilateral triangle, circumcenter and incenter are coincident.

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

$\begin{array}{rcl} S i n c e Point (1, 1) lies on the circle. \end{array}$

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

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

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

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