Q233 / 270 · NDA 2022 (Phase 1) · Mathematics

What is the equation of the circle which touches both the axes in the first quadrant and the line $y-2=0$ ?

  • A
    $x^2+y^2-2 x-2 y-1=0$
  • B
    $x^2+y^2+2 x+2 y+1=0$
  • Correct answer
    C
    $x^2+y^2-2 x-2 y+1=0$
  • D
    $x^2+y^2-4 x-4 y+4=0$

Solution

Since circle touches both the axes in the first quadrant and the line $y-2=0$ $\therefore$ centre $(1,1)$ radius $=1$ $\therefore$ Equation of circle is $\begin{aligned} & (\mathrm{x}-1)^2+(\mathrm{y}-1)^2=1 \\ & \Rightarrow \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-2 \mathrm{y}+1=0\end{aligned}$

Hence, the correct answer is option: 'C'.

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