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(D) The equation of the circle is $x^2 + y^2 + 10x + 12y + c = 0$.The center is $(-5, -6)$ and the radius $R = \sqrt{(-5)^2 + (-6)^2 - c} = \sqrt{25 +

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$25$

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(D) The equation of the circle is $x^2 + y^2 + 10x + 12y + c = 0$.The center is $(-5, -6)$ and the radius $R = \sqrt{(-5)^2 + (-6)^2 - c} = \sqrt{25 + 36 - c} = \sqrt{61 - c}$.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 $27\sqrt{3} \text{ sq. units}$,then $c$ is equal to

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