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