UPSC NDA 2 2021 Mathematics Solved Paper

Concept:The circumcentre is the intersection point of the perpendicular bisectors of the sides of a triangle.Explanation:The vertices of the triangle are found from the given graph.$A = (2, 1)$, $B = (2, -1)$, $C = (6, -1)$.The midpoint of $BC$ is $M = \left( \frac{2+6}{2}, \frac{-1-1}{2} \right) = (4, -1)$.The midpoint of $AC$ is $N = \left( \frac{2+6}{2}, \frac{-1+1}{2} \right) = (4, 0)$.Side $BC$ is horizontal, so its perpendicular bisector $MO$ is vertical through $M$: equation $x = 4$.Slope of $AC$ is $\frac{-1-1}{6-2} = \frac{-2}{4} = -\frac{1}{2}$.Thus slope of $NO$ (perpendicular to $AC$) is $2$.Equation of $NO$ through $N(4,0)$: $y - 0 = 2(x - 4) \implies y = 2x - 8$.Intersect $x=4$ with $y=2x-8$: $x=4$, $y=2(4)-8=0$.Hence the circumcentre is $(4,0)$.Answer:$(4, 0)$ (Option A).