Q93 / 160 · TS EAMCET 2007 · Mathematics

The equation of the circle of radius 3 that lies in the fourth quadrant and touching the lines $x=0$ and $y=0$ is

  • Correct answer
    A
    $x^2+y^2-6 x+6 y+9=0$
  • B
    $x^2+y^2-6 x-6 y+9=0$
  • C
    $x^2+y^2+6 x-6 y+9=0$
  • D
    $x^2+y^2+6 x+6 y+9=0$

Solution

Given, radius $=3$, Since, the circle touching both the coordinate axes in 4th quadrant, so equation is $\begin{gathered}\quad(x-3)^2+(y+3)^2=3^2 \\ \Rightarrow \quad x^2+9-6 x+y^2+9+6 y=9 \\ \Rightarrow \quad x^2+y^2-6 x+6 y+9=0\end{gathered}$

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

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