The radius of the larger circle lying in the | vedclass

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(B) Let the circle have radius $r$. Since the circle lies in the first quadrant and touches both coordinate axes,its center is $(r, r)$.The equation o

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$6$

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(B) Let the circle have radius $r$. Since the circle lies in the first quadrant and touches both coordinate axes,its center is $(r, r)$.The equation of the circle is $(x - r)^2 + (y - r)^2 = r^2$.The circle touches the line $4x + 3y - 12 = 0$. The perpendicular distance from the center $(r, r)$ to the line must be equal to the radius $r$:$\frac{|4r + 3r - 12|}{\sqrt{4^2 + 3^2}} = r$$\frac{|7r - 12|}{5} = r$$|7r - 12| = 5r$This gives two cases:$1) 7r - 12 = 5r$ $\Rightarrow 2r = 12$ $\Rightarrow r = 6$$2) 7r - 12 = -5r$ $\Rightarrow 12r = 12$ $\Rightarrow r = 1$The question asks for the radius of the larger circle,which is $r = 6$.

The radius of the larger circle lying in the first quadrant and touching the line $4x + 3y - 12 = 0$ and the coordinate axes is:

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