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Question

The equation of the circumcircle of an equilateral triangle is x2+y2+2gx+2fy+c=0 and one vertex of the triangle is 1,1 . The equation of incircle of the triangle is


A

4x2+y2=g2+f2

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B

4x2+y2+8gx+8fy=1-g1+3g+1-f1+3f

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C

4x2+y2+8gx+8fy=g2+f2

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D

None of these

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Solution

The correct option is B

4x2+y2+8gx+8fy=1-g1+3g+1-f1+3f


Explanation for the correct answer:

Finding equation of incircle of the triangle:

Given that the equation of circumcircle in an equilateral triangle is x2+y2+2gx+2fy+c=0 .

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

The circumcentre from the equation x2+y2+2gx+2fy+c=0 is -g,-f and so 1,1 will also pass through circumcircle.

Hence,

1+1+2g+2f+c=0c=-2g+f+1

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

rincircle=12g2+f2-c

So, equation of incircle with radius rincircle=12g2+f2-c and with center -g,-f is given by

⇒x+g2+y+f2=14g2+f2-c

⇒x2+y2+2gx+2fy+g2+f2=14g2+f2+12g+f+1

⇒4x2+y2+8gx+8fy=1-g1+3g+1-f1+3f

Hence, option (B) is the correct answer.


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