The incentre of an equilateral triangle is (1,1) and the equation of ene side is $3 x+4 y+3=0 .$ Then, the equation of the circumeircle of the triangle is

• A

  $x^{2}+y^{2}-2 x-2 y-2=0$
• Correct answer
  B

  $x^{2}+y^{2}-2 x-2 y-14=0$
• C

  $x^{2}+y^{2}-2 x-2 y+2=0$
• D

  $x^{2}+y^{2}-2 x-2 y+14=0$

Solution

since, triangle is equilateral the refore incentre (1,1) lies on the centroid of the $\Delta A B C$. $\therefore G D=$ Length of perpendicular from the point $G(1,1)$ to the line $3 x+4 y+3=0$ $A G=2 G D=4$ $\therefore$ Equation of circumcircle with centre at (1,1) $$ \begin{array}{l} \text { and radius = } \begin{aligned} 4 \text { units } \\ (x-1)^{2}+(y-1)^{2} &=4^{2} \\ \Rightarrow \quad x^{2}+y^{2}-2 x-2 y-14 &=0 \end{aligned} \end{array} $$

Hence, the correct answer is option: 'B'.