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The equation of the circumcircle of an equilateral triangle is `x^2+y^2+2gx+2fy+c=0`and one vertex of the triangle in (1, 1). The equation of the incircleof the triangle is`4(x^2+y^2)=g^2+f^2``4(x^2+y^2)=8gx+8fy=(1-g)(1+3g)+(1-f)(1+3f)``4(x^2+y^2)=8gx+8fy=g^2+f^2``non eoft h e s e`A. `4(x^(2)+y^(2))=g^(2)+f^(2)`B. `4(x^(2)+y^(2))+8gx+8fy=(1-g)(1+3g)+(1-f)(1+3f)`C. `4(x^(2)+y^(2))+8gx+8fy=g^(2)+f^(2)`D. none of these |
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Answer» Correct Answer - 2 In an equilateral triangle, circumcenter and incenter are coincident. Therefore, Inceter`-= ( -g,-f)` Point (1,1) lies on the circle. Therefore, `1^(2)+1^(2)+2g+2f+c=0` or `c= -2(g+f+1)` Also, in an equilateral triangle, Circumradius`=2xx` Inradius `:. `Inradius `=(1)/(2) xx sqrt(g^(2)+f^(2)-c)` Therefore, the equation of the incircle is `(x+g)^(2)+(y+f)^(2)=(1)/(4)(g^(2)+f^(2)-c)` ,brgt `=(1)/(4){g^(2)+f^(2)+2(g+f+1}` |
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