Question

If $2x-4y=9$ and $6x-12y+7=0$ are Parallel tangents to the circle, then radius of the circle is
(a) $\dfrac{\sqrt{3}}{5}$
(b) $\dfrac{17}{6\sqrt{5}}$
(c) $\dfrac{\sqrt{2}}{3}$
(d) $\dfrac{17}{3\sqrt{5}}$

Answer 1

Hint: Firstly find out the slope and intercepts of the two given equations. We know the diameter of the circle is equal to the distance between two parallel tangent lines to solve it. Try to represent and interpret graphically.

Complete step-by-step answer:
Now, let’s consider the one of the given tangents equation,
$2x-4y=9$
$\Rightarrow 4y=2x-9$
$\Rightarrow y=\dfrac{x}{2}-\dfrac{9}{4}$
This is of the form $y=mx+c$ where m is the slope.
Here, slope of the this tangent is $\dfrac{1}{2}$ and intercept is ${{c}_{1}}=-\dfrac{9}{4}$
Now, consider the other tangent equation,
$\begin{aligned} & 6x-12y+7=0 \\\ & 12y=6x+7 \\\ &\Rightarrow y=\dfrac{6}{12}x+\dfrac{7}{12} \\\ \end{aligned}$
$\Rightarrow y=\dfrac{x}{2}+\dfrac{7}{12}$
This is also of the form $y=mx+c$ where m is the slope.
Slope of this tangent is $\dfrac{1}{2}$ and intercept ${{c}_{2}}=\dfrac{7}{12}$
If slopes are equal then the lines are parallel to each other. So the slopes of the given tangents are equal, hence, the given two tangents are parallel to each other. It can be represented as shown below.

So, the diameter of the circle will be the distance between the two parallel tangents to the circle.
We know, the distance between 2 parallel lines $y=mx+{{c}_{1}}$ and $y=mx+{{c}_{2}}$ is given by
$d=\dfrac{\left| {{c}_{1}}-{{c}_{2}} \right|}{\sqrt{\left( 1+{{m}^{2}} \right)}}$
Where ${{c}_{1}}$,${{c}_{2}}$are intercepts of the two lines and $m$is the slope of the parallel lines.
Now substituting the values obtained from our calculations, i.e.,
${{c}_{2}}=\dfrac{7}{12}$,${{c}_{1}}=-\dfrac{9}{4}$ and $m=\dfrac{1}{2}$ in the distance formula, we get
$\begin{aligned} & d=\dfrac{\left| \dfrac{7}{12}+\dfrac{9}{4} \right|}{\sqrt{\left( 1+{{\left( \dfrac{1}{2} \right)}^{2}} \right)}} \\\ & d=\dfrac{\left| \dfrac{7}{12}+\dfrac{9}{4} \right|}{\sqrt{\left( 1+\dfrac{1}{4} \right)}} \\\ \end{aligned}$
Taking the LCM in both numerator and denominator, we get
$\begin{aligned} & d=\dfrac{\left| \dfrac{7+9\times 3}{12} \right|}{\sqrt{\left( \dfrac{4+1}{4} \right)}} \\\ & \Rightarrow d=\dfrac{\left| \dfrac{34}{12} \right|}{\sqrt{\left( \dfrac{5}{4} \right)}} \\\ & \Rightarrow d=\dfrac{34}{12}\times \dfrac{2}{\sqrt{5}} \\\ & \Rightarrow d=\dfrac{17}{3\sqrt{5}} \\\ \end{aligned}$
So, the diameter of the given circle is $\dfrac{17}{3\sqrt{5}}$ units.
Now we know, radius is half of diameter, so
$r=\dfrac{d}{2}$
Substituting the value of diameter, we get
$r=\dfrac{\dfrac{17}{3\sqrt{5}}}{2}=\dfrac{17}{6\sqrt{5}}$
Therefore, the radius of the given circle is $\dfrac{17}{6\sqrt{5}}$ units.
Hence, the correct answer is option (b).

Note: Students often make mistakes and forget to divide the distance between the tangents by to 2, to get the radius. So they will get the radius as $\dfrac{17}{3\sqrt{5}}$, this is the wrong answer. So they may select option (d) as the answer.