The equation of the circle which touches the | vedclass

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(A) Since the circle touches both coordinate axes and lies in the first quadrant,its center is $(c, c)$ and its radius is $r = c$.The distance from th

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$1$

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(A) Since the circle touches both coordinate axes and lies in the first quadrant,its center is $(c, c)$ and its radius is $r = c$.The distance from the center $(c, c)$ to the line $\frac{x}{3} + \frac{y}{4} = 1$ (or $4x + 3y - 12 = 0$) must be equal to the radius $c$.Using the perpendicular distance formula: $\left| \frac{4c + 3c - 12}{\sqrt{4^2 + 3^2}} \right| = c$.$\left| \frac{7c - 12}{5} \right| = c$.This gives two cases:Case $1$: $7c - 12 = 5c$ $\Rightarrow 2c = 12$ $\Rightarrow c = 6$.Case $2$: $7c - 12 = -5c$ $\Rightarrow 12c = 12$ $\Rightarrow c = 1$.Since the line $\frac{x}{3} + \frac{y}{4} = 1$ has intercepts $3$ and $4$,the circle must be smaller than the triangle formed by the axes and the line. For $c=6$,the circle lies outside the triangle. Thus,$c=1$ is the correct value for the circle touching the line internally.

The equation of the circle which touches the coordinate axes and the line $\frac{x}{3} + \frac{y}{4} = 1$,and whose centre lies in the first quadrant,is ${x^2} + {y^2} - 2cx - 2cy + {c^2} = 0$,where $c$ is:

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