The radius of a circle whose center lies in t | vedclass

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(C) Let the radius of the circle be $r$. Since the circle lies in the fourth quadrant and touches the lines $x=0$ and $y=0$,its center must be at $(r,

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Let the radius of the circle be $r$. Since the circle lies in the fourth quadrant and touches the lines $x=0$ and $y=0$,its center must be at $(r, -r)$,where $r > 0$.The distance from the center $(r, -r)$ to the line $3x+4y-12=0$ must be equal to the radius $r$.Using the perpendicular distance formula,we have:$\left| \frac{3(r) + 4(-r) - 12}{\sqrt{3^2 + 4^2}} \right| = r$$\left| \frac{3r - 4r - 12}{5} \right| = r$$\left| \frac{-r - 12}{5} \right| = r$$|r + 12| = 5r$Since $r > 0$,$r + 12 = 5r$ or $r + 12 = -5r$.Case $1$: $4r = 12 \Rightarrow r = 3$.Case $2$: $6r = -12 \Rightarrow r = -2$ (rejected as $r > 0$).Thus,the radius is $3$ units.

The radius of a circle whose center lies in the fourth quadrant and touches each of the three lines $x=0$,$y=0$,and $3x+4y-12=0$ is .... units.

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