Question

If $2x-4y=9$ and $6x-12y+7=0$ are common tangents to the circle, then radius of circle is

• A. $\frac{\surd 3}{5}$
• B. $\frac{17}{6\surd 5}$
• C. $\frac{\surd 2}{3}$
• D. $\frac{17}{3\surd 5}$

Answer 1

The correct option is B

$\frac{17}{6\surd 5}$

Explanation for the correct option:

Step 1. Find the radius of the circle:

Given lines are,

$2x-4y=9$

$\Rightarrow$ $4y=2x-9$

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

Step 2. Differentiate it with respect to $x$:

$\Rightarrow$ $\frac{dy}{dx}=\frac{1}{2}$

$\therefore m_1=\frac{1}{2}$

Now, second equation is

$6x-12y+7=0$

$\Rightarrow$ $y=\frac{x}{2}+\frac{7}{12}$

Step 3. Differentiate it with respect to $x$:

$\Rightarrow$$\frac{dy}{dx}=\frac{1}{2}$

$\therefore m_2=\frac{1}{2}$

$\because m_1=m_2$, then the lines are parallel.

$\because$The given lines are tangent to a circle then distant between them is diameter.

$\begin{array}{rcl}\therefore d& =& \frac{\left|b_2-b_1\right|}{\sqrt{1+m^2}}\\ & =& \frac{{\displaystyle \frac{7}{12}}+{\displaystyle \frac{9}{4}}}{\sqrt{1+{\displaystyle \frac{1}{4}}}}\\ & =& \frac{17+271}{12\times {\displaystyle \frac{\sqrt{5}}{2}}}\\ & =& \frac{34}{6\sqrt{5}}\\ & =& \frac{17}{3\sqrt{5}}\end{array}$

$\therefore$The required radius $=\frac{d}{2}$

$=\frac{\mathbf{17}}{\mathbf{6}\sqrt{\mathbf{5}}}$ units

Hence, Option ‘B’ is Correct.