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(B) In an equilateral triangle,the incentre,centroid,circumcentre,and orthocentre coincide. Let the incentre be $G(1, 1)$.The distance $r$ from the in

Vedclass · Accepted Answer

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

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(B) In an equilateral triangle,the incentre,centroid,circumcentre,and orthocentre coincide. Let the incentre be $G(1, 1)$.The distance $r$ from the incentre $G(1, 1)$ to the side $3x + 4y + 3 = 0$ is the inradius:$r = \frac{|3(1) + 4(1) + 3|}{\sqrt{3^{2} + 4^{2}}} = \frac{|3 + 4 + 3|}{\sqrt{9 + 16}} = \frac{10}{5} = 2$.In an equilateral triangle,the circumradius $R$ is twice the inradius $r$. Therefore,$R = 2r = 2(2) = 4$.The circumcircle has its centre at the incentre $(1, 1)$ and radius $R = 4$.The equation of the circle is $(x - 1)^{2} + (y - 1)^{2} = 4^{2}$.$x^{2} - 2x + 1 + y^{2} - 2y + 1 = 16$.$x^{2} + y^{2} - 2x - 2y + 2 = 16$.$x^{2} + y^{2} - 2x - 2y - 14 = 0$.

The incentre of an equilateral triangle is $(1, 1)$ and the equation of one side is $3x + 4y + 3 = 0$. Then,the equation of the circumcircle of the triangle is

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