Question

If the area of an equilateral triangle inscribed in the circle, $x^2+y^2+10x+12y+c=0$ is $27\sqrt{3}$ sq. units then $c$ is equal to :

• A. $13$
• B. $20$
• C. $25$
• D. $-25$

Answer 1

The correct option is C $25$

Let the side of an equiliateral triangle be $a$, radius of circle be $r$ and $O$ be the centre of circle.
Area of the equilateral triangle inscribed in the circle
$\frac{\sqrt{3}}{4}{a}^2=27\sqrt{3}\Rightarrow a=6\sqrt{3}$
As $\angle OAD={30}^{\circ },$
so $r\cos {30}^{\circ }=\frac{6\sqrt{3}}{2}\Rightarrow r=6$
$\therefore 6=\sqrt{25+36-c}$
$\Rightarrow c=25$