Q. 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 :
Solution:
$3 \left(\frac{1}{2} r^{2} .\sin120^{\circ}\right) =27 \sqrt{3}$
$ \frac{r^{2}}{2} \frac{\sqrt{3}}{2} = \frac{27\sqrt{3}}{3} $
$ r^{2} = \frac{108}{3} =36 $
Radius $ = \sqrt{25+36 -C}= \sqrt{36}$
$ C = 25 $
