The equation of the circle which touches both the axes and the line $\frac { x }{ 3 } +\frac { y }{ 4 } =1$ and lies in the first quadrant is $(x-c)^{ 2 }+(y-c)^{ 2 }={ c }^{ 2 }$ where c is ;

Given that, the circle touches both the axes and the line $\frac{x}{3}+\frac{y}{4}=1$ and lies in Qudrant-1.

Equation is $(x-c)^{2}+(y-c)^{2}=c^{2}$

It is of the form $(x-a)^{2}+(y-b)^{2}=r^{2}$ where $(a, b)$ is the centre of the circle and $r$ is its radius.

$\Rightarrow$ Centre of the given circle is $(c, c)$

$\text { Radius }=c$

It can be drawn as follows:

As the line touches the circle, it is tangent which will be perpendicular to the line joining the centre.

we know that if $\left(x_{1}, y_{1}\right)$ is joined to line $a x+b y+c=0$. then the length of perpendicular, $d=\frac{\left|a x_{1}+b y_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}$

$\begin{array}{l} \Rightarrow \quad c=\frac{|4 c+3 c-12|}{\sqrt{4^{2}+3^{2}}}=\frac{|7 c-12|}{5} \\ \Rightarrow \quad 5 c=7 c-12,5 c=-(7 c-12) \\ \Rightarrow 2 c=12,12 c=12 \\ \Rightarrow \quad c=6, c=1 \\ \therefore \quad c=1 \text { or } 6 \end{array}$