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 incircle of 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`

Options

A. `4(x^(2)+y^(2))=g^(2)+f^(2)`
B. `4(x^(2)+y^(2))+8gx+8fy=g^(2)+f^(2)`
C. `4(x^(2)+y^(2))+8gx+8fy=(1-g)(1+3g)+(1-f)(1+3f)`
D. none of these

Answer

C

Explanation

In an equilateral triangle the circumcentre and incentre coincide. So , the coordinates of the incentre are (-g, -f). Also, in an equilateral triangle, circumradius is twice the in-radius,
`:.` In-radius `=(1)/(2)` (Circum-radius)`=(1)/(2)sqrt(g^(2)+f^(2)-c)`
`:.` In-radius`=(1)/(2)sqrt(g^(2)+f^(2)+2+2g+2f)=(1)/(2)sqrt((g+1)^(2)+(f+1)^(2))`
Hence, the equation of the in-circle is
`(x+g)^(2)+(y+f)^(2)=(1)/(4)[(g+1)^(2)+(f+1)^(2)]`
`rArr 4(x^(2)+y^(2))+8 gx + 8fy=1+2g-3g^(2)+1+2f-3f^(2)`
`rArr 4 (x^(2)+y^(2))+8gx+8fy=(1-g)(1+3g)+(1-f)(1+3f)`.