Question

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

• A. $1.9$
• B. $0.95$
• C. $2.9$
• D. $1.45$

Answer 1

The correct option is B

$0.95$

Explanation for the correct Answer

Given,

$3x-4y+5=0$

$6x-8y-9=0$

We can rewrite it as

$3x-4y-\frac{9}{2}=0$

Here,

$a=3,b=4$

$c_1=5$ and $c_2=-\frac{9}{2}$

We know that, If two lines are parallel and touching the circle, then the radius of a circle is half the distance between the tangents.

Distance $d$ between two parallel lines $ax+by+c_1=0$ and$ax+by+c_2=0$ is given by

$d=\frac{\mid c_1-c_2|}{\sqrt{a^2+b^2}}$
$=\frac{\left|5+{\displaystyle \frac{9}{2}}\right|}{\sqrt{9+16}}$
​
$=\frac{19}{10}$

By the definition above the radius of a circle is half the distance between the tangents

So,
​∴ Radius $=\frac{1}{2}\times \frac{19}{10}$
$$\begin{gathered} =\frac{19}{20} \\ =0.95 \end{gathered}$$

Hence the radius of the circle touching the lines $3x-4y+5=0$and $6x-8y-9=0$ is $0.95$