The equations of the circles,which touch both | vedclass

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(B, C) Since the circle touches both axes in the first quadrant,its center is $(r, r)$ and its radius is $r$.The perpendicular distance from the cente

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$x^{2}+y^{2}-12x-12y+36=0$

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(B, C) Since the circle touches both axes in the first quadrant,its center is $(r, r)$ and its radius is $r$.The perpendicular distance from the center $(r, r)$ to the line $4x+3y-12=0$ is equal to the radius $r$.$\frac{|4r+3r-12|}{\sqrt{4^{2}+3^{2}}} = r$$\frac{|7r-12|}{5} = r$$|7r-12| = 5r$Case $1$: $7r-12 = 5r$ $\Rightarrow 2r = 12$ $\Rightarrow r = 6$.The equation is $(x-6)^{2}+(y-6)^{2} = 6^{2} \Rightarrow x^{2}+y^{2}-12x-12y+36=0$.Case $2$: $7r-12 = -5r$ $\Rightarrow 12r = 12$ $\Rightarrow r = 1$.The equation is $(x-1)^{2}+(y-1)^{2} = 1^{2} \Rightarrow x^{2}+y^{2}-2x-2y+1=0$.Thus,the possible equations are $x^{2}+y^{2}-2x-2y+1=0$ and $x^{2}+y^{2}-12x-12y+36=0$.

The equations of the circles,which touch both the axes and the line $4x+3y=12$ and have centers in the first quadrant,are

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