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