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# Draw two direct common tangents to two circles of radii 4cm and 2cm whose centres are 8cm apart. Measure the length of the tangents.  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. 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.
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