If the area of an equilateral triangle inscribed in the circle x2 + y2 + 10x + 12y + c = 0…

Question

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 :

Accepted Answer

$$3\left( {{1 \over 2}{r^2}.\sin {{120}^o}} \right) = 27\sqrt 3 $$

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

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

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

$C = 25$

Solutions · 1