
Draw two direct common tangents to two circles of radii 4cm and 2cm whose centres are 8cm apart. Measure the length of the tangents.
Answer
602.1k+ views
Hint:We will be using the concept of circle tangents to solve the problem. Also, some concepts of geometry and trigonometry will be used. Remember the tangent drawn to the circle is perpendicular to the radius at the point of contact.
Complete step-by-step answer:
Now, we have to find the length of direct tangents to the circles. We have been given that the radius of two circles are 4cm and 2cm respectively.
We will first draw a diagram to clear the situation.
We have been given that radius of circles are R = 4cm & r = 2cm………(1)
Also, the distance between the two centres is $ 8cm={{O}_{1}}{{O}_{2}}..........\left( 2 \right) $
Now, to find the length of AB, we will draw perpendicular from $ {{O}_{2}} $ to AO, as shown in the figure.
Now, in quadrilateral $ AO,{{O}_{2}}B $ all the angles are $ 90{}^\circ $ . Since, tangent is perpendicular to radius and O,F is perpendicular to AO. Therefore, $ AO,{{O}_{2}}B $ is a rectangle.
Now, In $ \Delta O,{{O}_{2}}E $ , we know that $ \angle {{O}_{2}}E{{O}_{1}}=90{}^\circ $ .
Therefore, by applying Pythagoras theorem,
$ \begin{align}
& {{O}_{1}}{{O}_{2}}^{2}=E{{O}_{1}}^{2}+E{{O}_{2}}^{2} \\
& {{O}_{1}}{{O}_{2}}^{2}={{\left( R-r \right)}^{2}}+E{{O}_{2}}^{2} \\
\end{align} $
Now using (1) and (2),
$ \begin{align}
& {{\left( 8 \right)}^{2}}={{\left( 4-2 \right)}^{2}}+{{\left( E{{O}_{2}} \right)}^{2}} \\
& 64=4+{{\left( E{{O}_{2}} \right)}^{2}} \\
& E{{O}_{2}}=\sqrt{64-4} \\
& E{{O}_{2}}=\sqrt{60} \\
& =2\sqrt{15}cm \\
\end{align} $
Now, $ AB=E{{O}_{2}} $ . Since, opposite sides of the rectangle are equal. So, $ AB=2\sqrt{15}cm $ .
Similarly, the length of the tangent CD will be equal to that of AB as the exact same process is required to find it and since the constants remain the same we will get the same answer therefore, the length of the direct common tangents is $ 2\sqrt{15}cm $ .
Note: These types of questions are usually tricky to do the first time but if one remembers the concepts of geometry it can be solved easily. Also we have to construct a perpendicular on our own to solve the question. This is a very important concept. properties of tangent drawn to the circle and pythagoras theorem is important here to get the solution.
Complete step-by-step answer:
Now, we have to find the length of direct tangents to the circles. We have been given that the radius of two circles are 4cm and 2cm respectively.
We will first draw a diagram to clear the situation.
We have been given that radius of circles are R = 4cm & r = 2cm………(1)
Also, the distance between the two centres is $ 8cm={{O}_{1}}{{O}_{2}}..........\left( 2 \right) $
Now, to find the length of AB, we will draw perpendicular from $ {{O}_{2}} $ to AO, as shown in the figure.
Now, in quadrilateral $ AO,{{O}_{2}}B $ all the angles are $ 90{}^\circ $ . Since, tangent is perpendicular to radius and O,F is perpendicular to AO. Therefore, $ AO,{{O}_{2}}B $ is a rectangle.
Now, In $ \Delta O,{{O}_{2}}E $ , we know that $ \angle {{O}_{2}}E{{O}_{1}}=90{}^\circ $ .
Therefore, by applying Pythagoras theorem,
$ \begin{align}
& {{O}_{1}}{{O}_{2}}^{2}=E{{O}_{1}}^{2}+E{{O}_{2}}^{2} \\
& {{O}_{1}}{{O}_{2}}^{2}={{\left( R-r \right)}^{2}}+E{{O}_{2}}^{2} \\
\end{align} $
Now using (1) and (2),
$ \begin{align}
& {{\left( 8 \right)}^{2}}={{\left( 4-2 \right)}^{2}}+{{\left( E{{O}_{2}} \right)}^{2}} \\
& 64=4+{{\left( E{{O}_{2}} \right)}^{2}} \\
& E{{O}_{2}}=\sqrt{64-4} \\
& E{{O}_{2}}=\sqrt{60} \\
& =2\sqrt{15}cm \\
\end{align} $
Now, $ AB=E{{O}_{2}} $ . Since, opposite sides of the rectangle are equal. So, $ AB=2\sqrt{15}cm $ .
Similarly, the length of the tangent CD will be equal to that of AB as the exact same process is required to find it and since the constants remain the same we will get the same answer therefore, the length of the direct common tangents is $ 2\sqrt{15}cm $ .
Note: These types of questions are usually tricky to do the first time but if one remembers the concepts of geometry it can be solved easily. Also we have to construct a perpendicular on our own to solve the question. This is a very important concept. properties of tangent drawn to the circle and pythagoras theorem is important here to get the solution.
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Trending doubts
What is the median of the first 10 natural numbers class 10 maths CBSE

The draft of the Preamble of the Indian Constitution class 10 social science CBSE

Which women's tennis player has 24 Grand Slam singles titles?

Who gave "Inqilab Zindabad" slogan?

Why is there a time difference of about 5 hours between class 10 social science CBSE

Who is the Brand Ambassador of Incredible India?

