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(C) Since the circle lies in the first quadrant and touches both coordinate axes,its center is $(c, c)$ and its radius is $c$,where $c > 0$.The equati

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$1$ or $6$

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(C) Since the circle lies in the first quadrant and touches both coordinate axes,its center is $(c, c)$ and its radius is $c$,where $c > 0$.The equation of the circle is $(x-c)^2 + (y-c)^2 = c^2$.The line is $\frac{x}{3} + \frac{y}{4} = 1$,which can be written as $4x + 3y - 12 = 0$.Since the circle touches this line,the perpendicular distance from the center $(c, c)$ to the line must be equal to the radius $c$.$\frac{|4c + 3c - 12|}{\sqrt{4^2 + 3^2}} = c$$\frac{|7c - 12|}{5} = c$$|7c - 12| = 5c$Case $1$: $7c - 12 = 5c \implies 2c = 12 \implies c = 6$.Case $2$: $7c - 12 = -5c \implies 12c = 12 \implies c = 1$.Thus,$c = 1$ or $c = 6$.

If the equation of the circle lying in the first quadrant,touching both the coordinate axes and the line $\frac{x}{3}+\frac{y}{4}=1$ is $(x-c)^2+(y-c)^2=c^2$,then $c=$

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