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(D) The given circle equation is $x^2+y^2+10x+12y+c=0$. Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=5$ and $f=6$. The radius

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$25$

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(D) The given circle equation is $x^2+y^2+10x+12y+c=0$. Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=5$ and $f=6$. The radius $R$ of the circle is given by $R = \sqrt{g^2+f^2-c} = \sqrt{25+36-c} = \sqrt{61-c}$. For an equilateral triangle inscribed in a circle of radius $R$,the side length $a$ is given by $a = R\sqrt{3}$. The area of the equilateral triangle is $\frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4}(R\sqrt{3})^2 = \frac{3\sqrt{3}}{4}R^2$. Given the area is $27\sqrt{3}$,we have $\frac{3\sqrt{3}}{4}R^2 = 27\sqrt{3}$. $R^2 = 27 	imes \frac{4}{3} = 36$. Since $R^2 = 61-c$,we have $61-c = 36$. $c = 61-36 = 25$.

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

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