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(A) Since the circle touches both coordinate axes in the first quadrant,its center is $(r, r)$ and its radius is $r$,where $r > 0$. The equation of th

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

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(A) Since the circle touches both coordinate axes in the first quadrant,its center is $(r, r)$ and its radius is $r$,where $r > 0$. The equation of the circle is $(x-r)^2 + (y-r)^2 = r^2$. The distance from the center $(r, r)$ to the line $4x+3y-6=0$ must be equal to the radius $r$. Thus,$\frac{|4r+3r-6|}{\sqrt{4^2+3^2}} = r$. $\frac{|7r-6|}{5} = r$. This gives two cases: $7r-6 = 5r$ or $7r-6 = -5r$. Case $1$: $2r = 6 \implies r = 3$. The center is $(3, 3)$ and the equation is $(x-3)^2 + (y-3)^2 = 3^2$,which simplifies to $x^2+y^2-6x-6y+9=0$. Case $2$: $12r = 6 \implies r = 1/2$. The center is $(1/2, 1/2)$ and the equation is $(x-1/2)^2 + (y-1/2)^2 = (1/2)^2$,which simplifies to $x^2+y^2-x-y+1/4=0$,or $4x^2+4y^2-4x-4y+1=0$. The line $4x+3y-6=0$ passes through $(1.5, 0)$ and $(0, 2)$. For the center $(3, 3)$,$4(3)+3(3)-6 = 15 > 0$. For the center $(1/2, 1/2)$,$4(1/2)+3(1/2)-6 = 2+1.5-6 = -2.5 Since the circle lies below the line $L=0$ (meaning the center satisfies $4x+3y-6 Thus,the equation is $4x^2+4y^2-4x-4y+1=0$.

$A$ circle touches both the coordinate axes and the straight line $L \equiv 4x+3y-6=0$ in the first quadrant. If this circle lies below the line $L=0$,then the equation of that circle is

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