Question

If \[2x - 4y = 9\] and \[6x - 12y + 7 = 0\] are common tangents to the circle. Then find the radius of the circle.
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: In the given question, two equations of tangents to the circle are given. By differentiating the given equations with respect to \[x\], we will find the slope of the tangents. Then using the distance formula, we will find the diameter of the circle. To calculate the radius, divide the diameter by 2.
Formula Used:
The distance between two parallel lines is: \[d = \dfrac{{\left| {{c_1} - {c_2}} \right|}}{{\sqrt {1 + {m^2}} }}\]
   Where \[{c_1}\] and \[{c_2}\] are the constant of equation, and \[m\] is the slope of the parallel lines
\[radius = \dfrac{{diameter}}{2}\]
Complete step by step solution:
The given equations of the tangents to the circle are \[2x - 4y = 9\] and \[6x - 12y + 7 = 0\].
Let’s simplify the equations of the tangents.
\[2x - 4y = 9\]
\[ \Rightarrow \]\[4y = 2x - 9\]
\[ \Rightarrow \]\[y = \dfrac{1}{2}x - \dfrac{9}{4}\]
\[6x - 12y + 7 = 0\]
\[ \Rightarrow \]\[12y = 6x + 7\]
\[ \Rightarrow \]\[y = \dfrac{1}{2}x + \dfrac{7}{{12}}\]

Let’s differentiate the above equations with respect to \[x\] and find the slope of the tangents.
The slope of the tangent \[y = \dfrac{1}{2}x - \dfrac{9}{4}\] is:
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{1}{2}\]

The slope of the tangent \[y = \dfrac{1}{2}x + \dfrac{7}{{12}}\] is:
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{1}{2}\]

Since the slopes of both lines are equal. So, the lines are parallel to each other.
The given lines are tangents to the circle, and then the distance between them is the diameter of the circle.
Now apply the distance formula \[d = \dfrac{{\left| {{c_1} - {c_2}} \right|}}{{\sqrt {1 + {m^2}} }}\], to calculate the diameter of the circle.
Let \[d\] be the diameter of the circle.
\[d = \dfrac{{\left| {\dfrac{{ - 9}}{4} - \dfrac{7}{{12}}} \right|}}{{\sqrt {1 + {{\left( {\dfrac{1}{2}} \right)}^2}} }}\]
\[ \Rightarrow \]\[d = \dfrac{{\left| {\dfrac{{ - 108 - 28}}{{48}}} \right|}}{{\sqrt {1 + \dfrac{1}{4}} }}\]
\[ \Rightarrow \]\[d = \dfrac{{\left| {\dfrac{{ - 136}}{{48}}} \right|}}{{\sqrt {\dfrac{5}{4}} }}\]
\[ \Rightarrow \]\[d = \dfrac{{\left| {\dfrac{{ - 17}}{6}} \right|}}{{\dfrac{{\sqrt 5 }}{2}}}\]
\[ \Rightarrow \]\[d = \dfrac{{17}}{6} \cdot \dfrac{2}{{\sqrt 5 }}\]
\[ \Rightarrow \]\[d = \dfrac{{17}}{{3\sqrt 5 }}\]

Therefore, the radius of circle is,
required radius \[ = \dfrac{d}{2}\]
\[ \Rightarrow \]required radius \[ = \dfrac{{\dfrac{{17}}{{3\sqrt 5 }}}}{2}\]
\[ \Rightarrow \]required radius \[ = \dfrac{{17}}{{3\sqrt 5 }} \cdot \dfrac{1}{2}\]
\[ \Rightarrow \]required radius \[ = \dfrac{{17}}{{6\sqrt 5 }}\]
Hence the correct option is option B
Note: The distance formula \[d = \dfrac{{\left| {{c_1} - {c_2}} \right|}}{{\sqrt {1 + {m^2}} }}\] is used to calculate the shortest distance between two parallel lines. This formula is used when we know the constant terms and slope of the lines.