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 (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``

Options

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

Answer

2

Explanation

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}`