Question

The equilateral $\Delta ABC$ has vertices $B(1,0)$ and $C(5,0).$ If $A$ lies in the fourth quadrant, then the equation of the incircle of $\Delta ABC$ is

• A. $(x-3{)}^2+{(y-\frac{2\sqrt{3}}{3})}^2=1$
• B. $(x-3{)}^2+{(y-\frac{2\sqrt{3}}{3})}^2=\frac{4}{3}$
• C. $(x+3{)}^2+{(y-\frac{2\sqrt{3}}{3})}^2=\frac{4}{3}$
• D. $(x-3{)}^2+{(y-\frac{2\sqrt{3}}{3})}^2=\frac{4}{3}$

Answer 1

The correct option is D $(x-3{)}^2+{(y-\frac{2\sqrt{3}}{3})}^2=\frac{4}{3}$

Let the coordinate of $A$ is $(x,y)$. $A$ lies on the perpendicular bisector of the line $BC$. $\Rightarrow x=3$
$\sqrt{(1-3{)}^2+y^2}=4\Rightarrow y=-2\sqrt{3}$
$\therefore$ Centre of the incircle is $(3,\frac{-2\sqrt{3}}{3})$
Equation of the circle: $(x-3{)}^2+(y+\frac{2\sqrt{3}}{3}{)}^2=\frac{4}{3}$