Let the equation of the circle is
$$
x^2+y^2+2 g x+2 f y+c=0
$$
This circle touch the coordinate axes and lying in the first quadrant, then
$$
\begin{aligned}
g^2-c & =0 \text { and } f^2-c=0 \\
g & = \pm \sqrt{c}, f= \pm \sqrt{c}
\end{aligned}
$$
circle lies in first quadrant
$\therefore$ The centre is $(\sqrt{c}, \sqrt{c})$
If the line $4 x+3 y-12=0$ touch the circle, then
$$
\begin{aligned}
& \sqrt{f^2+g^2-c}=\frac{4 \sqrt{c}+3 \sqrt{c}-12}{\sqrt{4^2+3^2}} \\
& \Rightarrow \quad \sqrt{c+c-c}=\frac{7 \sqrt{c}-12}{5} \\
& \Rightarrow \quad 5 \sqrt{c}=7 \sqrt{c}-12 \\
& \Rightarrow \quad 2 \sqrt{c}=12 \Rightarrow c=36 \\
& \text { Radius }=\sqrt{f^2+g^2-c}=\sqrt{c+c-c}=\sqrt{c}=6
\end{aligned}
$$