The equation of the circle which touches the both axes and St. lines  4x  +3y - 6 = 0 in the first quadrant lies below if

$4x+3y-6=0\text{ touches the circle }\frac{7a-6}{\sqrt{16+9}}\pm a$

$\Rightarrow \mathrm{a}=3,\frac{1}{2}$

$\Rightarrow \mathrm{a}=\frac{1}{2}\text{ only because }(0,0)\text{ and }\left(\frac{1}{2},\frac{1}{2}\right)\text{ same side of line }$

$\text{ Required circle }={\left(x-\frac{1}{2}\right)}^2+{\left(y-\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^2$

$4x^2+4y^2-4x-4y+1=0$