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

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

$4 (x^{2} + y^{2}) + 8 (g^{2} + f^{2}) = 0$

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

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

answer is B.

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Detailed Solution

Incircle and circumcircle have same centre but radius is half of that of circumcircle
Also c = - (2g + 2f + 2) as (1, 1) lies on circumcircle.

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