WBJEE 2012 Math Solved Paper

Since, triangle is equilateral therefore incentre (1,1) lies on the centroid of the ΔABC.
∴ GD = Length of perpendicular from the point
G(1,1) to the line 3x+4y+3=0 3x+4y+3=0
=‌

3(1)+4(1)+3

√32+42

=2
AG = 2 GD = 4
∴ -.Equation of circumcircle with centre at (1,1) and radius = 4 units
(x−1)2+(y−1)2=42
⇒‌x2+y2−2x−2y−14=0