Answer
452.7k+ 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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)