UPSC NDA 2 2021 Mathematics Solved Paper

Concept:The center of a circle is equidistant from all points on its circumference, so the distances from the center to any three given points are equal.Explanation:Let the center be $O(x, y)$.Given three points: $A(5, -8)$, $B(-2, 9)$, and $C(2, 1)$.Set $OA^2 = OB^2$ and $OA^2 = OC^2$.From $OA^2 = OB^2$:$(x - 5)^2 + (y + 8)^2 = (x + 2)^2 + (y - 9)^2$Simplify: $x^2 - 10x + 25 + y^2 + 16y + 64 = x^2 + 4x + 4 + y^2 - 18y + 81$Cancel $x^2, y^2$: $-10x + 16y + 89 = 4x - 18y + 85$Rearrange: $-10x - 4x + 16y + 18y + 89 - 85 = 0$$-14x + 34y + 4 = 0$ → $14x - 34y = 4$ …(1)From $OA^2 = OC^2$:$(x - 5)^2 + (y + 8)^2 = (x - 2)^2 + (y - 1)^2$Simplify: $x^2 - 10x + 25 + y^2 + 16y + 64 = x^2 - 4x + 4 + y^2 - 2y + 1$Cancel $x^2, y^2$: $-10x + 16y + 89 = -4x - 2y + 5$Rearrange: $-10x + 4x + 16y + 2y + 89 - 5 = 0$$-6x + 18y + 84 = 0$ → $6x - 18y = 84$ …(2)Solve (1) and (2):From (2): $x = 3y + 14$ (dividing by 6).Substitute into (1): $14(3y + 14) - 34y = 4$ → $42y + 196 - 34y = 4$ → $8y = -192$ → $y = -24$Then $x = 3(-24) + 14 = -72 + 14 = -58$.Thus, the center is $(-58, -24)$.Answer:$(-58, -24)$ (Option D)