Question

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

• A. $4\left(x^2+y^2\right)=g^2+f^2$
• B. $4\left(x^2+y^2\right)+8gx+8fy=\left(1-g\right)\left(1+3g\right)+\left(1-f\right)\left(1+3f\right)$
• C. $4\left(x^2+y^2\right)+8gx+8fy=g^2+f^2$
• D. None of these

Answer 1

The correct option is B

$4\left(x^2+y^2\right)+8gx+8fy=\left(1-g\right)\left(1+3g\right)+\left(1-f\right)\left(1+3f\right)$

Explanation for the correct answer:

Finding equation of incircle of the triangle:

Given that the equation of circumcircle in an equilateral triangle is $x^2+y^2+2gx+2fy+c=0$ .

We know that for an equilateral triangle the circumcentre and the incenter are the same.

The circumcentre from the equation $x^2+y^2+2gx+2fy+c=0$ is $\left(-g,-f\right)$ and so $\left(1,1\right)$ will also pass through circumcircle.

Hence,

$$\begin{gathered} 1+1+2g+2f+c=0 \\ c=-2\left(g+f+1\right) \end{gathered}$$

Now, for an equilateral triangle the inradius is half of the circumradius which will be

$r_{\text{incircle}}=\frac{1}{2}\left(\sqrt{g^2+f^2-c}\right)$

So, equation of incircle with radius $r_{\text{incircle}}=\frac{1}{2}\left(\sqrt{g^2+f^2-c}\right)$ and with center $\left(-g,-f\right)$ is given by

$\Rightarrow {\left(x+g\right)}^2+{\left(y+f\right)}^2=\frac{1}{4}\left(g^2+f^2-c\right)$

$\Rightarrow x^2+y^2+2gx+2fy+g^2+f^2=\frac{1}{4}\left(g^2+f^2\right)+\frac{1}{2}\left(g+f+1\right)$

$\Rightarrow 4\left(x^2+y^2\right)+8gx+8fy=\left(1-g\right)\left(1+3g\right)+\left(1-f\right)\left(1+3f\right)$

Hence, option (B) is the correct answer.