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  • The equation of the circumcircle of an equilateral triangle is <img alt="\mathrm{x^2+y^2+2 g x+2 f y+c=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2&

The equation of the circumcircle of an equilateral triangle is \mathrm{x^2+y^2+2 g x+2 f y+c=0} and one vertex of the triangle is \mathrm{(1,1)}. The equation of incircle of the triangle is

Option: 1

\mathrm{4\left(x^2+y^2\right)=8^2+f^2}


Option: 2

\mathrm{4\left(x^2+y^2\right)+8 g x+8 f y=(1-g)(1+3 g)+(1+3 f)(1-f)}


Option: 3

\mathrm{4\left(x^2+y^2\right)+8 g x+8 f y=g^2+f^2}


Option: 4

\mathrm{\text { None of these }}


Answers (1)

best_answer

In an equilateral triangle, the circumcentre and the incentre are the same point.

\mathrm{\therefore \quad \text { incentre }=(-g,-f)}

\mathrm{\text { Also } 1^2+1^2+2 g+2 f+c=0}

Also, in an equilateral triangle, circumradius \mathrm{ =2 \times} inradius. 

\mathrm{ \therefore \quad \text { inradius }=\frac{1}{2} \times \sqrt{g^2+f^2-c} \text {. }}

\therefore   the equation of the incircle is 

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

Simplify

Posted by

manish

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