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

20

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25

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13

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- 25

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Solution

The correct option is B $25$
$3 (\frac{1}{2} r^{2} . sin 120^{o}) = 27 \sqrt{3}$

$\frac{r^{2}}{2} \frac{\sqrt{3}}{2} = \frac{27 \sqrt{3}}{3}$

$r^{2} = \frac{108}{3} = 36$

Radius $= \sqrt{25 + 36 - C} = \sqrt{36}$

$C = 25$
$∴$ Option (2)

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