The circle touching the coordinate axes with its centre lying on $x-2 y-3=0$ is
The circle touching the coordinate axes with its centre lying on $x-2 y-3=0$ is
$x^2+y^2-2 x+2 y+1=0$
$x^2+y^2+2 x-2 y+1=0$
$x^2+y^2+6 x+6 y-9=0$
$x^2+y^2-6 x-6 y+9=0$
Solution
The circle touches the co-ordinate axes and distance of circle
(radius) from axes is same
If centre is $(a, b)$ and radius $r$
$\begin{aligned} & |a|=|b|=r \\ & (x-a)^2+(y-b)^2=r^2 \\ & (a, b) \text { satisfies } x-2 y-3=0\end{aligned}$
$\begin{aligned} & a-2 b-3=0 \\ & a=2 b+3 \\ & (x-2 b-3)^2+(y-b)^2=r^2 \\ & |b|=r \Rightarrow r^2=b^2 \\ & (x-2 b-3)^2+(y-b)^2=b^2\end{aligned}$
circle pass through $(0, b)$
$\begin{aligned} & (0-2 b-3)^2+(b-b)^2=b^2 \\ & (2 b+3)^2=b^2 \\ & 2 b+3= \pm b \\ & b=-3 \text { and } b=-1 \\ & a=2 b+3 \\ & a=-3 \text { and } a=1\end{aligned}$
for $(a, b) \equiv(1,-1)$
Equation of circle is
$x^2+y^2-2 x+2 y+1=0$