If area of an equilateral triangle inscribed in the circle `x^2+y^2+10x+12y+c=0` is `27sqrt3`, then the value of c is

Options

A. 20
B. -25
C. 13
D. 25

Answer

D

Explanation

Clearly, centre of the circumscribed circle is the centroid (G) of the equilateral triangle ABC.
[`therefore` in an equilateral triangle circumference and centroid coincide]

Also, we know that
`DeltaAGB cong Delta BCG cong DeltaCGA` [by SAS cogruence rule]
`therefore ar(DeltaABC)=3ar(DeltaAGB)`
`=3((1)/(2)r^(2) sin 120^(@))`
`[therefore " area of triangle " =(1)/(2) ab sinZC)]`
`therefore ar (DeltaABC)=27sqrt3` [given]
`therefore (3)/(2)r^(2)(sqrt3)/(2)=27sqrt3`
`[sin120^(@)=sin(180^(@)-60^(@))=sin60^(@)=(sqrt3)/(2)]`
`rArr r^(2) = 4 xx 9`
`rArr r = 6`
Now, radius of circle,
`r = sqrt(g^(2)+f^(2)-c)`
`rArr 6 = sqrt(25+36-c)`
[`therefore` in the given equation of circle 2g = 10 and 2f = 12`rArr g = 5 and f =6]`
`rArr 36 = 25 + 36`
`rArr c= 25`