Question

The equation of 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 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

Answer 1

The correct option is B $4(x^2+y^2)+8gx+8fy=(1+g)(1-3g)+(1+f)(1-3f)$

Centre of the given circumcircle is $C(-g,-f)$
and circum-radius is $R=\sqrt{(g-1{)}^2+(f-1{)}^2}$, since one vertex of the triangle is given $(1,1)$ .
Now in equilateral we know inradius $r=\frac{R}{2}$ and the centers of both the circles coincide.
Hence equation of incircle is given by,
$(x+g{)}^2+(y+f{)}^2=r^2=\frac{R^2}{4}=\frac{1}{4}\left[\right(g-1{)}^2+(f-1{)}^2]$
$\Rightarrow 4(x^2+y^2)+8gx+8fy=(1-2g-3g^2)+(1-2f-3f^2)$
$\Rightarrow 4(x^2+y^2)+8gx+8fy=(1+g)(1-3g)+(1+f)(1-3f)$