The equation of the circumcircle of an equilateral triangle is $x^2+y^2+2gx+2fy+c=0$ and one vertex of the

triangle is$(1, 1)$. The equation of the incircle of the triangle, is

 

detailed solution

Correct option is C

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 =12 (Circum- radius) =12 82+f2−cBut, the circumcircle passes through (1, 1). Therefore1+1+2g+2f+c=0⇒c=−2−2g−2f∴ In-radius =12g2+f2+2+2g+2f=12(g+1)2+(f+1)2Hence, the equation of the in-circle is(x+g)2+(y+f)2=14(g+1)2+(f+1)2⇒ 4x2+y2+8gx+8fy=1+2g−3g2+1+2f−3f2⇒ 4x2+y2+8gx+8fy=(1−g)(1+3g)+(1−f)(1+3f)