If the incenter of an equilateral triangle is | vedclass

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(A) For an equilateral triangle,the incenter and circumcenter coincide at the same point $(1, 1)$.The inradius $r$ is the perpendicular distance from

Vedclass · Accepted Answer

$x^2 + y^2 - 2x - 2y - 14 = 0$

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(A) For an equilateral triangle,the incenter and circumcenter coincide at the same point $(1, 1)$.The inradius $r$ is the perpendicular distance from the incenter $(1, 1)$ to the side $3x + 4y + 3 = 0$:$r = \frac{|3(1) + 4(1) + 3|}{\sqrt{3^2 + 4^2}} = \frac{|3 + 4 + 3|}{5} = \frac{10}{5} = 2$.In an equilateral triangle,the circumradius $R$ is twice the inradius $r$:$R = 2r = 2(2) = 4$.The equation of the circumcircle with center $(1, 1)$ and radius $R = 4$ is:$(x - 1)^2 + (y - 1)^2 = 4^2$$x^2 - 2x + 1 + y^2 - 2y + 1 = 16$$x^2 + y^2 - 2x - 2y - 14 = 0$.

If the incenter of an equilateral triangle is $(1, 1)$ and the equation of one of its sides is $3x + 4y + 3 = 0$,then the equation of the circumcircle of this triangle is

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