Question

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

• A. $x^{2}+y^{2}-2x-2y-2=0$
• B. $x^{2}+y^{2}-2x-2y-14=0$
• C. $x^{2}+y^{2}-2x-2y+2=0$
• D. $x^{2}+y^{2}-2x-2y+14=0$

Answer 1

Answer: (B)

Since, triangle is equilateral therefore incentre (1, 1) lies on the centroid of the $\triangle A B C$.
$\therefore G D=$ Length of perpendicular from the point
$G(1,1)$ to the line $3 x+4 y+3=0$

$3 x+4 y+3=0$
$=\frac{3(1)+4(1)+3}{\sqrt{3^{2}+4^{2}}}=2$
$A G=2 G D=4$
$\therefore $ Equation of circumcircle with centre at $(1,1)$ and radius $=4$ units
$(x-1)^{2}+(y-1)^{2} =4^{2}$
$\Rightarrow x^{2}+ y^{2}-2 x-2 y-14 =0$