JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 21)
If the area of an equilateral triangle inscribed in the circle x2 + y2
+ 10x + 12y + c = 0 is $$27\sqrt 3 $$ sq units then c is equal to :
20
25
$$-$$ 25
13
Explanation
_10th_January_Evening_Slot_en_21_1.png)
$$3\left( {{1 \over 2}{r^2}.\sin {{120}^o}} \right) = 27\sqrt 3 $$
$${{{r^2}} \over 2}{{\sqrt 3 } \over 2} = {{27\sqrt 3 } \over 3}$$
$${r^2} = {{108} \over 3} = 36$$
Radius $$ = \sqrt {25 + 36 - C} = \sqrt {36} $$
$$C = 25$$
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