Find the area of an equilateral triangle inscribed in the circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .

Open in App

Solution

Given circle $x^{2} + y^{2} + 2 y x + 2 f y + c = 0$
Let $^{'} o^{'}$ center two $△ A B C$ in equilateral triangle
$o = [- 9, - f] O A = O B = O C = \sqrt{g^{2} + f^{2} - c}$
In $△^{l e} O B M = sin 60^{o} = \frac{B M}{O B}$
$B M = O △ sin 60^{o} = O B \times \frac{\sqrt{3}}{2}$
$∴ B C = 2 B M = \sqrt{3} O B$
Area of $△^{l e} A B C = \frac{\sqrt{3}}{4} (B C)^{7}$
$= \sqrt{\frac{3}{4}} \times 3 (O B)^{2}$
$= \frac{3 \sqrt{3}}{4} (g^{2} + f^{2} - c) s q . m t r$

Suggest Corrections

Similar questions

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

x^{2} + y^{2} - 6 x + 2 y - 28 = 0.

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

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 x = 0

Join BYJU'S Learning Program