An equilateral triangle inscribed in the circle x 2- y 2-2 g x -2 fy - c -0 then its side isA- 3 -g2-f2-cB- -3g2-f2-cC- -g2-f2-c-3D- 2 -g2-f2-c

The correct option is B r=√(g^2+f^2 c) AD^2=a^2 a^2/4 AD=√(3)/2a r=2/3AD =a/√(3) a=√(3)r =√(3(g^2+f^2 c)) ...

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Question

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

$3\sqrt{g^2+f^2-c}$

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$\sqrt{3(g^2+f^2-c)}$

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$\frac{\sqrt{g^2+f^2-c}}{\sqrt{3}}$

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

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Solution

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

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(c > 0)

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

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

The centre and radius of the circle $x^{2} + y^{2} + 2 g x + 2 y f + c = 0$ are and respectively.

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