UPSC NDA II 2018 Mathematics Solved Paper

Concept:When a circle touches the $y$-axis, the radius equals the absolute value of the $x$-coordinate of its centre.The normal at the point of tangency passes through the centre and meets the circle at the opposite end of the horizontal diameter.Explanation:The given circle is $x^{2} + y^{2} + gx + fy + \frac{e}{4} = 0$.Its centre is $C\left(-\frac{g}{2}, -\frac{f}{2}\right)$ and radius $R = \sqrt{\left(\frac{g}{2}\right)^2 + \left(\frac{f}{2}\right)^2 - \frac{e}{4}}$.Since the $y$-axis ($x = 0$) touches the circle, the distance from centre to $y$-axis equals the radius: $\left|-\frac{g}{2}\right| = R$.Thus the radius is $\frac{|g|}{2}$, and the point of tangency $A$ is $\left(0, -\frac{f}{2}\right)$ (on the $y$-axis at the same $y$-coordinate as the centre).The normal at $A$ passes through the centre $C$ and continues to the opposite point $B$ on the circle.Because the radius $CA$ is horizontal (from $\left(-\frac{g}{2},-\frac{f}{2}\right)$ to $\left(0,-\frac{f}{2}\right)$), the line $CA$ is the horizontal diameter.Going from $C$ to $A$ is a step of $+\frac{g}{2}$ in $x$ (assuming $g>0$; sign handled by absolute value).The opposite point $B$ is reached by a step of $-\frac{g}{2}$ from $C$: $x = -\frac{g}{2} - \frac{g}{2} = -g$, and $y = -\frac{f}{2}$ (unchanged).Hence the coordinates of $B$ are $\left(-g, -\frac{f}{2}\right)$.Answer:Option B: $\left(-g,-\frac{f}{2}\right)$