Question

The incentre of an equilateral triangle is $(1,1)$ and the equation of one side is $3x+4y+3=0$ 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 $△ABC$.
$\therefore GD=$ Length of perpendicular from the point
$G(1,1)$ to the line $3x+4y+3=0$

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