Question

Find the circumcentre of the triangle whose vertices are $(0,0)$, $(3,\sqrt{3})$ and $(0,2\sqrt{3})$.

• A. $(1,\sqrt{3})$
• B. $(-1,\sqrt{3})$
• C. $(0,\sqrt{3})$
• D. $(1,2\sqrt{3})$

Answer 1

The correct option is A $(1,\sqrt{3})$

Given the three vertices of a triangle

$A(3,\sqrt{3}),B(0,0)\text{and}C(0,2\sqrt{3})$

Mid point of $AB=(\frac{3+0}{2},\frac{\sqrt{3+0}}{2})=(\frac{3}{2},\frac{\sqrt{3}}{2})$

$\therefore$ Slope of $AB=\frac{y_2-y_1}{x_2-x_1}=\frac{0-\sqrt{3}}{0-3}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$

$\therefore$ Slope of the perpendicular bisector $=-\sqrt{3}$

$\therefore$ Equation of AB with slope $-\sqrt{3}$ and coordinates $(\frac{3}{2},\frac{\sqrt{3}}{2})$

$y-y_1=m(x-x_1)\Rightarrow y-\frac{\sqrt{3}}{2}=-\sqrt{3}(x-\frac{3}{2})\frac{2y-\sqrt{3}}{2}=\frac{-\sqrt{3}(2x-3)}{2}\Rightarrow 2y-\sqrt{3}=-2\sqrt{3}x+3\sqrt{3}2\sqrt{3x}+2y-4\sqrt{3}=0\to \left(1\right)$

Similarly for AC, $A(3,\sqrt{3})\text{and}A(3,\sqrt{3})$

Mid point of $AC=(\frac{3+0}{2},\frac{\sqrt{3}+2\sqrt{3}}{2})=(\frac{3}{2},\frac{3\sqrt{3}}{2})$

$\therefore$ Slope of $AC=\frac{y_2-y_1}{x_2-x_1}=\frac{2\sqrt{3}-\sqrt{3}}{0-3}=\frac{-\sqrt{3}}{3}=\frac{-1}{\sqrt{3}}$

$\therefore$ Slope of perpendicular bisector $=\sqrt{3}$

Equation of AC with slope $\sqrt{3}$ and the coordinates $(\frac{3}{2},\frac{3\sqrt{3}}{2})$

$y-y_1=m(x-x_1)\Rightarrow y-\frac{3\sqrt{3}}{2}=\sqrt{3}(x-\frac{3}{2})2y-3\sqrt{3}=2\sqrt{3x}-3\sqrt{3}\Rightarrow 2\sqrt{3x}-2y=0\to \left(2\right)\frac{2\sqrt{3x}+2y=4\sqrt{3}2\sqrt{3x}\pm 2y=0}{4y=4\sqrt{3}}\Rightarrow y=\sqrt{3}$

$\therefore x=1$

$\therefore$ Circumcenter of triangle$=(1,\sqrt{3})$