The equation of the 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 in circle 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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Solution

The correct option is B $4 (x^{2} + y^{2}) + 8 g x + 8 f y = (1 - g) (1 + 3 g) + (1 - f) (1 + 3 f)$
$(x + q)^{2} + (y + f)^{2} = g^{2} + f^{2} - C$
radius of circum circle = $\sqrt{g^{2} + f^{2} - C}$
$∴$ radius of incircle = $\frac{\sqrt{g^{2} + f^{2} - C}}{2}$
$∴$ equation of incircle
$(x + g)^{2} + (y + f)^{2} = \frac{g^{2} + f^{2} - C}{4}$
(1, 1) satisfies
$x^{2} + y^{2} + 2 g x + 2 f y + C = 0$
$∴ C = - 1 - 2 g - 2 f$
$∴$ Equation of incircle
$4 (x + g)^{2} + 4 (x + f)^{2} = g^{2} + f^{2} + 2 + 2 g + 2 f$
$4 (x^{2} + y^{2}) + 8 g + 8 f y = 1 + 2 g - {3 g}^{2} + 1 + 2 f + {3 f}^{2} = (1 - g) (1 + 3 g) (1 + 3 f) (1 - f)$

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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

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

(1, 1)

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 $\sqrt{g^{2} + f^{2} + c}$ ?

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

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

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