Question:
If the area of an equilateral triangle inscribed in the circle, $x^{2}+y^{2}+10 x+12 y+c=0$ is $27 \sqrt{3}$ sq.
units then $\mathrm{c}$ is equal to :
Correct Option: , 2
Solution:
$3\left(\frac{1}{2} \mathrm{r}^{2} \cdot \sin 120^{\circ}\right)=27 \sqrt{3}$
$\frac{r^{2}}{2} \frac{\sqrt{3}}{2}=\frac{27 \sqrt{3}}{3}$
$r^{2}=\frac{108}{3}=36$
Radius $=\sqrt{25+36-C}=\sqrt{36}$
$\therefore$ Option (2)
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