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

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Aicharya · Accepted Answer

Given lines,

$\begin{aligned} & l_1: x=0 \\ & l_2: y=0\end{aligned}$ $l_3: 3 x+4 y-12=0$ $\Rightarrow \quad y=\frac{-3}{4} x+3$ $\because$ Circle touches $X$-axis and $Y$-axis. $\Rightarrow$ Then, perpendicular distance from $X$-axis and $Y$-axis to center is same and let's say it $r$. $\therefore \text { Center }=(r, r)$ Let circle touches $l_3$ at $P$. $\therefore \quad O P=r$ $\left|\frac{3 r+4 r-12}{\sqrt{3^2+4^2}}\right|=r$ $\Rightarrow \quad \pm r=\frac{7 r-12}{5}$ When $r=\frac{7 r-12}{5}$ $\begin{aligned} \Rightarrow & & 2 r & =12 \\ \Rightarrow & & r & =6\end{aligned}$ and when $-r=\frac{7 r-12}{5}$ $\Rightarrow \quad 12 r=12$ $r=1$

The radius of a circle whose center lies in the fourth quadrant and touches each of the three lines $x=0,…

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