# 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) $\dfrac{17}{3\sqrt{5}}$

Last updated date: 19th Mar 2023

•

Total views: 305.7k

•

Views today: 7.84k

Answer

Verified

305.7k+ views

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{align}

& 6x-12y+7=0 \\

& 12y=6x+7 \\

&\Rightarrow y=\dfrac{6}{12}x+\dfrac{7}{12} \\

\end{align}$

\[\Rightarrow y=\dfrac{x}{2}+\dfrac{7}{12}\]

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{align}

& 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{align}$

Taking the LCM in both numerator and denominator, we get

$\begin{align}

& 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{align}$

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.

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{align}

& 6x-12y+7=0 \\

& 12y=6x+7 \\

&\Rightarrow y=\dfrac{6}{12}x+\dfrac{7}{12} \\

\end{align}$

\[\Rightarrow y=\dfrac{x}{2}+\dfrac{7}{12}\]

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{align}

& 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{align}$

Taking the LCM in both numerator and denominator, we get

$\begin{align}

& 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{align}$

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.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE