Question

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

• A. $x^2+y^2-2x-2y-2=0$
• B. $x^2+y^2-2x-2y+2=0$
• C. $x^2+y^2-2x-2y-7=0$
• D. $x^2+y^2-2x-2y-14=0$

Answer 1

Answer: (D)

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

$r\left(inradius\right)=\frac{|3.1+4.1+3|}{\sqrt{3^2+4^2}}=2$
As for equilateral triangle circumcentre and incentre coincide.
$sin30=\frac{2}{R}$
$R=4$
Equation of circle with centre(1,1) and radius $=4$
${(x-1)}^2+{(y-1)}^2=4^2=16$
$x^2+y^2-2x-2y-14=0$