The radius of any circle touching the lines 3 | vedclass

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(D) The given lines are $L_1: 3x - 4y + 5 = 0$ and $L_2: 6x - 8y - 9 = 0$. We can rewrite $L_2$ as $3x - 4y - \frac{9}{2} = 0$. Since the lines are pa

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$\frac{19}{20}$

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(D) The given lines are $L_1: 3x - 4y + 5 = 0$ and $L_2: 6x - 8y - 9 = 0$. We can rewrite $L_2$ as $3x - 4y - \frac{9}{2} = 0$. Since the lines are parallel,the distance between them is equal to the diameter of the circle. The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$. Here,$d = \frac{|5 - (-9/2)|}{\sqrt{3^2 + (-4)^2}} = \frac{|5 + 4.5|}{5} = \frac{9.5}{5} = \frac{19/2}{5} = \frac{19}{10}$. Since the diameter $2r = d$,we have $2r = \frac{19}{10}$,which implies $r = \frac{19}{20}$. Thus,the correct option is $D$.

The radius of any circle touching the lines $3x - 4y + 5 = 0$ and $6x - 8y - 9 = 0$ is

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