UPSC NDA 2 2021 Mathematics Solved Paper

Concept:The general equation of a circle is $x^{2} + y^{2} + 2gx + 2fy + c = 0$, and its radius is $\sqrt{g^{2} + f^{2} - c}$.Explanation:Let the circle be $x^{2} + y^{2} + 2gx + 2fy + c = 0$.It passes through $(5, -8)$, $(-2, 9)$, and $(2, 1)$.Substitute each point:$5^{2} + (-8)^{2} + 2g(5) + 2f(-8) + c = 0$ gives $10g - 16f + c = -89$. (1)$(-2)^{2} + 9^{2} + 2g(-2) + 2f(9) + c = 0$ gives $-4g + 18f + c = -85$. (2)$2^{2} + 1^{2} + 2g(2) + 2f(1) + c = 0$ gives $4g + 2f + c = -5$. (3)Subtract (2) from (1): $14g - 34f = -4$ → $7g - 17f = -2$. (4)Subtract (3) from (2): $-8g + 16f = -80$ → $g - 2f = 10$. (5)Solve (4) and (5): From (5), $g = 2f + 10$. Substitute into (4): $7(2f + 10) - 17f = -2$ → $14f + 70 - 17f = -2$ → $-3f = -72$ → $f = 24$.Then $g = 2(24) + 10 = 58$.Use (3): $4(58) + 2(24) + c = -5$ → $232 + 48 + c = -5$ → $c = -285$.Radius $r = \sqrt{g^{2} + f^{2} - c} = \sqrt{58^{2} + 24^{2} - (-285)} = \sqrt{3364 + 576 + 285} = \sqrt{4225} = 65$.Thus $r = 65 > 60$.Answer:Option D: $r > 60$ is correct.