No. of circles which touch both the Axes and whose centre lies on $\mathrm{x}-2\mathrm{y}=3$ is

Let the centre (a,b) & Radius r

Since the circle touches the coordinates axes the distance of the centre from the axes will be same. Thus, |a|=|b|=r

${(x-a)}^2+{(y-b)}^2=r^2$

(a,b) satisfies the equation x−2y=3 then a−2b=3

a=2b+3

${(x-2b-3)}^2+{(y-b)}^2=r^2$

Since |b|=r

${(x-2b-3)}^2+{(y-b)}^2=b^2$

(0,b) is on the circle then

$$\begin{gathered} {(0-2b-3)}^2+{(b-b)}^2=b^2 \\ \Rightarrow b^2+4b+3=0 \end{gathered}$$

$\Rightarrow$b=−1,−3 

then a=1,−3

Hence, the centre of the circles are (1,−1) & (−3,−3)