UPSC NDA I 2022 Mathematics Solved Paper

Concept:A circle that touches both axes in the first quadrant has its center at $(r, r)$ and radius $r$.It also touches a given horizontal line; equate the distance from the center to that line with $r$ to find $r$.Explanation:The circle touches the $x$-axis and $y$-axis in the first quadrant. So its center is $(r, r)$, where $r$ is the radius.The given line is $y - 2 = 0$, i.e., $y = 2$.The distance from the center $(r, r)$ to the line $y = 2$ is $\left| r - 2 \right|$.Since the circle touches the line, this distance must equal the radius $r$.Thus $\left| r - 2 \right| = r$.In the first quadrant, $r > 0$, so $2 - r = r$ (since $r < 2$ for the circle to be below the line).Solving gives $2 = 2r$, hence $r = 1$.Therefore the center is $(1, 1)$ and radius $1$.The equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$.Substituting $h = 1$, $k = 1$, $r = 1$ gives:$(x - 1)^2 + (y - 1)^2 = 1^2$.Expand: $x^2 - 2x + 1 + y^2 - 2y + 1 = 1$.Simplify: $x^2 + y^2 - 2x - 2y + 2 = 1$.Thus $x^2 + y^2 - 2x - 2y + 1 = 0$.Answer:The correct equation is $x^{2} + y^{2} - 2x - 2y + 1 = 0$, which matches option C.