If the area of an equilateral triangle inscribed in the circle x^2+y^2+10x+12y+c=0 is 27√3…

Question

If the area of an equilateral triangle inscribed in the circle $x^2+y^2+10x+12y+c=0$ is $27\sqrt3$ sq. units, then $c$ is equal to:

Accepted Answer

Centre $(-5,-6)$, $r^2=61-c$. Equilateral triangle inscribed in a circle of radius $r$ has area $\dfrac{3\sqrt3}4r^2$:
$$\frac{3\sqrt3}4r^2=27\sqrt3\Rightarrow r^2=36\Rightarrow61-c=36\Rightarrow c=25.$$

Solutions · 1