Question

If the centroid of an equilateral triangle is $(1,1)$and its one vertex is $(-1,2)$, then the equation of its circumcircle is:

• A. $x^2+y^2-2x-2y-3=0$
• B. ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}-2\mathrm{y}-3=0$
• C. $x^2+y^2+2x+2y-3=0$
• D. none of these

Answer 1

The correct option is A

$x^2+y^2-2x-2y-3=0$

Explanation for the correct option.

Given the centroid of the equilateral triangle is $(1,1)$ & its vertex is $(-1,2)$.

We know that in an equilateral triangle, the centroid, circumcenter and the incenter are at the same point.
So, the radius of the circumcircle is the distance between $(1,1)$ and $(-1,2)$.

$\begin{array}{rcl}r& =& \sqrt{{\left(-1-1\right)}^2+{\left(2-1\right)}^2}\\ & =& \sqrt{5}\end{array}$
So, the equation of circumcircle is:

$\begin{array}{rcl}(x-1{)}^2+(y-1{)}^2& =& 5\\ \therefore x^2+y^2-2x-2y-3& =& 0\end{array}$
Hence, option A is correct.