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 ?

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

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

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

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

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Solution

The correct option is C $4 (x^{2} + y^{2}) + 8 g x + 8 f y = (1 - g) (1 + 3 g) + (1 - f) (1 + 3 f)$
Center $= (- g, - f)$ vertex $= (1, 1)$
Radius $R^{2} = (1 + g)^{2} + (1 + f)^{2}$
In equilateral triangle
radius of incircle $= \frac{1}{2}$ radius of circumcircle
$r = \frac{1}{2} R$
Equation of circle
$4 (x + g)^{2} + 4 (y + f)^{2} = (1 + g)^{2} + (1 + f)^{2} 4 (x^{2} + y^{2}) + 8 g x + 8 f y = (1 - g) (1 + 3 g) + (1 - f) (1 + 3 f)$

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Similar questions

An equilateral triangle inscribed in the circle $x^{2} + y^{2} + 2 g x + 2 f y + c$ = 0 then its side is

g^{2} + f^{2} = c,

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

Which of the following equations represents a circle with centre $(g, f)$ and radius $\sqrt{g^{2} + f^{2} + c}$ ?

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

g^{2} < c < f^{2}

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

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