Question

The radius of the circle touching the straight lines $x-2y-1=0$ and $3x-6y+7=0$ is

• A. $\frac{3}{\sqrt{5}}$
• B. $\frac{\sqrt{5}}{3}$
• C. $\sqrt{5}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

Answer: (B)

$\frac{\sqrt{5}}{3}$

Diameter of circle=distance of the point (1,0)
from $3x-6y+7=0$
$\therefore$ $\frac{3\left(1\right)-6\left(0\right)+7}{\sqrt{{\left(3\right)}^2+{(-6)}^2}}=\frac{10}{\sqrt{45}}=\frac{2}{3}\sqrt{5}$
Now, radius of circle $=\frac{1}{2}\left(\frac{2}{3}\sqrt{5}\right)=\frac{\sqrt{5}}{3}$