The radius of any circle touching the lines $3 x-4 y+5=0,6 x-8 y-9=0$ is
The radius of any circle touching the lines $3 x-4 y+5=0,6 x-8 y-9=0$ is
1
$\frac{23}{15}$
$\frac{20}{19}$
$\frac{19}{20}$
Solution
Given lines are $3 x-4 y+5=0$
$\Rightarrow$
$
\begin{aligned}
& 6 x-8 y-9=0 \\
& 3 x-4 y-\frac{9}{2}=0
\end{aligned}
$
Given lines (i) and (ii) are parallel to each other and they are touching the required circle
$\therefore$ Distance between lines (i) and (ii) = diameter of circle
$
\begin{gathered}
\frac{\left|C_1-C_2\right|}{\sqrt{a^2+b^2}}=2 r \\
\frac{\left|5+\frac{9}{2}\right|}{\sqrt{3^2+(-4)^2}}=2 r \\
\therefore \quad 2 r=\frac{19 / 2}{5} \Rightarrow 2 r=\frac{19}{10} \Rightarrow r=\frac{19}{20}
\end{gathered}
$
Hence, option (4) is correct