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(A) The given equation of the incircle is $x^2 + y^2 + 4x - 6y + 4 = 0$. Its center (incenter) is $(-2, 3)$ and its inradius $r$ is $\sqrt{(-2)^2 + (3

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$x^2 + y^2 + 4x - 6y - 23 = 0$

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(A) The given equation of the incircle is $x^2 + y^2 + 4x - 6y + 4 = 0$. Its center (incenter) is $(-2, 3)$ and its inradius $r$ is $\sqrt{(-2)^2 + (3)^2 - 4} = \sqrt{4 + 9 - 4} = \sqrt{9} = 3$. In an equilateral triangle,the incenter and the circumcenter coincide. Therefore,the circumcenter is $(-2, 3)$. Also,in an equilateral triangle,the circumradius $R$ is twice the inradius $r$,so $R = 2r = 2 	imes 3 = 6$. The equation of the circumcircle is $(x - (-2))^2 + (y - 3)^2 = R^2$. $(x + 2)^2 + (y - 3)^2 = 6^2$. $x^2 + 4x + 4 + y^2 - 6y + 9 = 36$. $x^2 + y^2 + 4x - 6y + 13 - 36 = 0$. $x^2 + y^2 + 4x - 6y - 23 = 0$.

If the equation of the incircle of an equilateral triangle is $x^2 + y^2 + 4x - 6y + 4 = 0$,then the equation of its circumcircle is:

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